I think there's a simpler solution. Let: x = y³ Square both sides: 1 + √(1+y³) = y² Minus 1 and square: 1 + y³ = y⁴ - 2y² + 1 Rearrange and factor: 0 = y⁴ - y³ - 2y² Factorise: 0 = y²(y² - y - 2) Factorise: 0 = y²(y - 2)(y + 1) Substitute: x = 0, x = 8, x = -1 Check: x = 8 is the only solution.
@@beng4186 I'm talking about the multivariable roots. Let √x=1. So x²=1 and x can be +1 or -1 as -1² also equals 1. We consider square roots generally as +ve so most of the time we ignore -1. Haven't you heard of the cube roots of unity?
interestingly its only one solution, x=8, and no complex once for fun tested √(1+√(1±x) )= (±x)¹/ ³ ..the solutions is 8, -8, 1, -1 ... ∛is cubroot symbol (the YT font does a poor job displaying it clearly), using ¹/ ³ instead
I used a substitution for √(1 + x). This makes x = m^2 - 1. Sub in and raise both sides to the 6th power. Since m + 1 =\= 0, divide both sides by (m + 1)^2. Simple after that. -1 and 8, with -1 extraneous
It's way simpler without utilizing the difference of squares or sum of cubes. The term 1 on both sides just cancels out, then you can factorize the polynomial and get smth like u^2(u^2-u-2). The quadratic is easily factorizable by guess to (u-2)(u+1).
Same for me. I solved it and then watched the video wondering the whole time WHY was he making it so complicated?!? (Probably to throw in the formulas as a bonus lesson, I guess...)
The reason you cannot do what you suggest is that your solution requires you to divide both sides by u + 1. There is no guarantee that u -1, which results in dividing by zero.
@@salkabalani1482 no. If you just move everything to one side, you can factor without any division. And even if there was a division, we would just have to check u=1 separately.
Instead of any substitution, I did it by raising LHS and RHS to the power of 6 (LCM of 2 and 3), then ending with an equation containing only √(x+1) and other x terms, which then can be squared to get a polynomial equation, and solve them... It takes some steps to solve, but hey, its a solution...
If someone aa curious as me, there are always some ways to make extraneous solutions work. x=-1 is solution if we assume external square root as a branch of multivalued function fixed by sqrt(1) = -1 x=0 is solution if internal square root is a branch fixed by the same thing as above.
x = t²+2t LHS = √(1+t+1) = √(t+2) RHS = ³√[t(t+2)] Raising both sides to the 6th power, (t+2)³=t²(t+2)² => (t+2)²(t²-t-2) = 0 and hence t = -2, -1 or 2 => x = 0, -1 or 8 A simple check shows x=8 is the only solution.
I loved the geometric representation at the beginning of the explanation. Makes it so much easier to visualize where these algebraic equations originated.
Incredible. Another great one. I had to pause the video at points but I understood the method. I'm glad you mentioned not to stop since after that long calculation I would have chosen all three numbers to be the final answer which is wrong.
It's easier than that: squaring both sides you get (1 + x)^(1/2) + 1 = x^(2/3). Since the product of (1 + x)^(1/2) + 1 and (1 + x)^(1/2) - 1 is x, we have that (1 + x)^(1/2) - 1 = x^(1/3). Subtracting the two equations gives x^(2/3) - x^(1/3) = 2. Letting y = x^(1/3) we get that y^2 - y - 2 = 0, which has solutions y = 2 and y = -1, corresponding to x = 8 and x = -1 respectively. Only the first fits the original equation, so x = 8 is the only solution.
"Since the product of (1 + x)^(1/2) + 1 and (1 + x)^(1/2) - 1 is x, we have that (1 + x)^(1/2) - 1 = x^(1/3)" How did you get to that from that product?
@@maxhagenauer24fooled me for a moment too: Since the first equation gave us 1+sqrt(1+x) = x^2/3 [1+sqrt(1+x)] * x^1/3 = x Now noting that [sqrt(1+x)+1] [sqrt(1+x)-1] = x Setting them equal we can divide by the like term and get x^1/3 = sqrt(1+x) -1
I did that too, but it didn't clearly mean it was the only solution without a little analysis -- I looked at neighborhoods of the solution to convince myself, it was the only one.
I didn't even have to use substitution After squaring both sides I got: 1+√(1+x)=³√x² √(1+x)=³√x² -1 Squaring both sides again 1+x=(³√x² -1)² 1+x=³√x⁴ -2•³√x² +1 x=³√x⁴ -2•³√x² Dividing both sides by ³√x² x^(1-2/3)=³√x² -2 ³√x=³√x² -2 Rearranging the terms to resemble a quadratic (³√x)² -³√x -2=0 Factoring (³√x -2)(³√x +1)=0 And I get ³√x=2 and ³√x=-1 So x=8,-1, but after checking them the only viable solution is x=8
To eliminate both square and cube roots, take the 6th power of both sides of the equation. Expanding and rearranging terms gives: (4+x)√(1+x) = x²-3x-4 = (x+1)(x-4). Dividing both sides by √(1+x) (since x≠-1) gives 4+x = (x-4)√(1+x). Squaring again, expanding and collecting like terms gives: x³-8x² = 0, which has the obvious solution x=8. (The extraneous solutions x=0 and x=-1 are solutions to similar equations with √ replaced by -√.)
Nice! But i got an even simpler solution, in which we dont have to deal with sum and difference of cubes, which inadvertently leads us to making a calculation error somewhere most of the time. If we square the original equation (we will get x^2/3, but we don't have to worry about those weird exponents), take the 1 on the other side, and square it again to remove the root inside the root, then we get an equation in x with fractional powers (the equation is : x^4/3 - 2x^2/3 - x) The equation sure looks scary but it can be simplified quite nicely by taking a substitution of cuberoot (x) = t. After the substitution, we can factor in and solve the equation hence obtained extremely easily, and get t=0,0,2,-1 which subsequently gives us x = 0,0,8,-1 and checking back we can eliminate the redundant solutions and get x = 8 as a solution. EASY RIGHT?
I did the calculation in the old-fashioned way, entirely without substitution. I raised both sides to the sixth power. (You can also first square both sides and then cube them both, if you prefer that.) Then, I isolated the term with the root and squared the equation again. One might assume that this results in a super unwieldy equation but after simplifying the terms, one is left with this: 7x³ + 8x² = x⁴ We can further simplify by dividing by x² (and testing whether x²=0 and therefore x=0 is a solution; it is of course not one), and we get this simple quadratic equation: x² - 7x - 8 = 0 This equation has two solutions, x=-1 and x=8; testing shows that x=-1 is an extraneous solution but x=8 is valid. And that's it! 😀
@kintagrama What I meant is the following: We start with the equation as seen: √[1+√(1+x)] = ∛x Then, raising both sides to the 6th power (because we need to eliminate the square root on the left side and the cube root on the right side) gives us: [1+√(1+x)]³ = x² Expanding the cubed sum on the left hand side results in: 1³ + 3⋅1²⋅√(1+x) + 3⋅1⋅√(1+x)² + √(1+x)³ = x² And now comes the isolation I was talking about. We got rid of the root on the right side in our very first step but there are still roots in the second and fourth addend. (The root in the third addend is canceled by the square.) So we can rearrange the equation in a way that we have all addends with a root on one side and all others on the other side: 3 √(1+x) + √(1+x)³ = x² - 3 (1+x) - 1 Next, we can rewrite √(1+x)³ as √(1+x)² ⋅ √(1+x)¹ - and do a little clean-up on the right side. The root and square of the first factor again cancel out, leaving us with the same root we already have in the first addend: 3 √(1+x) + (1+x) √(1+x) = x² - 3x - 4 In a final step, we can now factor out the root on the left side: (x+4) √(1+x) = x² - 3x - 4 And that's it! This is what I meant by isolating the term with the root. The idea is that we can now square both sides without this in resulting in another root anywhere - something you can't do as long as you have a root in a sum. (For example, if you have 4+√x on one side, you can square this all day long but you'll always get another root: (4+√x)² = 4²+8√x+x. The first and third summand aren't roots but the middle one is. If you then group the root and non-root terms together (x+16)+(8√x) and square another, you'll again get two addends without roots but the middle one keeps the root. This will never change unless you can move all non-root addends on the other side.) Does all of this make sense?
Just from the expected shapes of the functions I would only expect one solution if there is any. You have a cube root and something that will generally act like a fourth root. Those two only intersect at 0 and 1, but the adding of 1s will shift it around so I don’t expect them to meet up at both of those two spots anymore. Not a solution, but the quick analysis so I can see if what I get in the end makes sense.
Do the u substitution and rearrange till you get -u^4 + u^3 + 2u^2 = 0, factor out u^2 and solve for u to be 0, 2, and -1. Sub these back in to find x and you get that the only one that works is 8.
I raised both to the sixth power, then eventually got to: (x+4)(x+1)^0.5 = x^2-3x-4 Squared both sides of that and got: x^4-7x^3-8x^2=0 This factors to (x-8)(x+1)(x^2)=0. The rest follows the same way as 8:55 on.
It is possible to dispense with the extraneous solutions at the very start by noticing that the left-hand side of the original equation can't have a value of less than 1 for every possible value of x in the domain of that formula. This means that cbrt(x) can't be less than 1; that is, x can't be less than 1. Since x=8 is the only solution that fits this criterion, it's the only solution of the equation.
@asr2009 has the right of it. In general, extraneous solutions are introduced by the math used to solve the problem. For radical equations, extraneous solutions can occur when you raise the equation to an even power (this is because doing so destroys information about signs). For a different type of extraneous solution, rational equations (where there is a variable in the denominator) can have extraneous solutions introduced by multiplying. When you plug those in to check, they generally cause division by zero. In all cases, extraneous solutions are solutions that do not work in the original equation, but only in the transformed one.
@@davidtripp2118 When I looked up the graphs of *√(1+√(1+x)) = x⅓* and it's transformed equation *x⁴ - 7x³ - 8x² = 0* on Desmos, It showed X = 8 for first equation and X = -1 & 8 for the second equation. We can clearly see that X = 0 is also a solution for the second equation, so why does it eliminate that possibility?
The extra solutions only "work" if you take the negative square roots of the original equation -- not something done in formal mathematics! For example, for x = 0 you can take the negative square root of the inner sqrt(1 + x) term to get -1, then add that to the 1 and get sqrt(0) = 0^1/3 = 0. Also for x = -1, the inner sqrt term is zero (1 + -1), resulting in sqrt(1) = (-1)^1/3. Take the negative sqrt of 1 there to get -1 = -1. It's issues like these that caused mathematicians to regulate using only the "principle square root" in equations, which is why the 0 and -1 solutions don't work in practice.
I did this with way less steps and not needing difference of cubes or squares. I just didn't do a u substitution and worked on getting rid of the squares, followed by some simplification and factoring. I was able to get the same answers in about about 9 lines of work. I do like that you can take advantage of the difference of cubes and squares if you want.
Thank goodness for UA-cam's speed control, as you go at rocket pace much too fast for me to follow along. Playing at 0.75x makes it actually possible to follow. If I miss one single step I'm screwed, that's why going so fast is a problem.
Answer 3 Explanation: Square both sides first Then Solve using substitute √1+x = t then substitute x= t^2 - 1 and make equation in 't' then solve for t quadratic equation then t=2 and x will equal to' 3' So 3 is the answer
Why not get to x^(4/3)-2x^(2/3)= x and substitute x=y^3. y^4-2y^2=y^3. Check for y=0 and then solve y^2-y=2. y=2 or -1 and x=8 or -1 but x=-1 is invalid.
The answer is 8, simply done by trying trying out numbers that gave an integer answer on the left. So sqrt(x+1) must be integer. X=0 doesn't work, as we get sqrt(2) on the left, but the next candidate x=3 gives 2 on the left and the right. Of course, if the answer wasn't integer, this wouldn't work, but if time was critical in a test, worth a few seconds to try it.
I solved it but I think I made it much harder for myself. I didn't substitute at all and only brute forced it by cubing and squaring the entire equation and distributing the terms whenever possible. At one point I had to distribute two terms both with 3 different terms and thus ended up with 9 terms. I was surprised that in the end I was left with a quadratic equation that had a really simple discriminant. And after checking the solutions, those being -1 and 8, 8 turned out to be the correct solution. I'd like to know if anyone else went about the problem this way or if it's just a needlessly complicated process
Easiest way is to look 1- (1+x)=1 .which we can use to first square the exp and than take reciprocal. 1. Square -> 1+ sqrt(1+x) = x^2/3 2 .take reciprocal-> 1-sqrt(1-x)=x^(1/3)=sqrt(1+sqrt(1+x)) 3. Now square and solve U get quadratic
I absolutely loved the visual explanation of the diff of squares and cubes! While I agree with many of the people here about the simpler way of doing it. I did the same all the while wondering if I was losing any mathematical rigor in the process (I have learned that you can’t cancel things willy nilly even tho’ it may be obvious). am curious to know why you did not go that route.
That is the most roundabout and incoherent solution video I've ever seen. This is what I did: Square both sides and rearrange to get: sqrt(1+x) = x^(2/3) - 1 Square both sides again to get: 1+x = x^(4/3) - 2x^(2/3) + 1 Simplification yields: x^(4/3) - 2x^(2/3) - x = 0 Now make the following substitution: u = x^1/3 u^4 - u^3 - 2u^2 = 0 It's obvious that you can factor out u^2 here. u^2 (u^2 - u - 2) = 0 u=0 is a root. Solving the quadratic in parenthesis yields: u = (-1,2) Now back-substitute to convert u's to x's. x = u^3 x = (0^3, (-1)^3, 2^3) = (0,-1,8) Check the above values for x against the original problem equation to determine actual solution(s) and discard any extraneous roots. When running that check, only one solution emerges: x = 8. Done.
I did this doing two substitutions, x = a^2-1, then a = b^2-1, then cubed both sides and solved to get b = -1,0,2 going back through the substitution chain, I get x = -1, 0, 8. -1 and 0 are extraneous, leaving 8.
To everyone complaining that he didn't use the most direct methodology to find the solution(s), you need to realize he's also teaching application of the sum and difference of squares in an easy problem so that people will know and remember to use it in more complex problems. This is a teaching channel. Don't forget that.
First, let’s set ( y = \sqrt[3]{x} ). This means ( y^3 = x ). Substituting ( y ) into the equation, we get: [ \sqrt[3]{1 + y} = y ] Cubing both sides to eliminate the cube root, we have: [ 1 + y = y^3 ] Rearranging the equation, we get: [ y^3 - y - 1 = 0 ] Now, we need to solve this cubic equation for ( y ). Using numerical methods or algebraic manipulation, we find that one of the roots of this equation is ( y = 2 ). Substituting ( y = 2 ) back into ( y^3 = x ), we get: [ 2^3 = x ] So, ( x = 8 ). Therefore, the correct solution to the equation is ( x = 8 ).
Hey, please read this, I got this problem stuck in my head for the past month. If I draw a rectangle and on the long side of it I draw a square with the same dimensions as the long side and then I lengthen the side to infinity, the area of the square is infinity but also the one of the rectangle but the squares infinity is bigger than the infinity of the rectangle, why?
As others have pointed out, this solution is not the simplest. Here's what I did: (1 + (1 + x)^(1/2))^(1/2) = x^(1/3) 1 + (1 + x)^(1/2) = x^(2/3) -Square both sides (1 + x)^(1/2) = x^(2/3) - 1 -Move 1 to the other side 1 + x = x^(4/3) - 2x^(2/3) + 1 -Square both sides x^(4/3) - x - 2x^(2/3) = 0 -Move everything to one side (1's cancel) [x^(2/3)][x^(2/3) - x^(1/3) - 2] = 0 -Factor out x^(2/3) [x^(2/3)][x^(1/3) - 2][x^(1/3) + 1] = 0 -Factor the quadratic in terms of x^(1/3) Then each factor can be set equal to 0 and solved for x. The same solutions of 0, 8, and -1 emerge. Only 8 works.
I solved this in a simpler manner, I think. I used the same u substitution and squared both sides, getting sqrt(1 - u^3) = u^2 - 1. I squared both sides again to get 1 + u^3 = u^4 - 2u^2 + 1. The ones cancel, and moving everything to the left we have u^4 - u^3 - 2u^2 = 0. This factors to (u^2)(u^2 - u - 2) = 0 Note that u cannot be negative, as the cube root of x is a principal square root. Also note that u cannot be 0 because we can test it in the original equation and show it will not work. Therefore u is strictly positive. Since u cannot be zero, we can divide both sides by u^2 with no worries, leaving is with the quadratic which is easy and yields u = 2 or -1. u is strictly positive, so u = 2, and since x = u^3, x = 8.
@@himanshuuu6361 I think you missed my point, I was saying you don’t know if 8 is the only solution I was not at all referring to how to guess a solution
Retrace your steps and do them in reverse, if there is no problem then all the solutions should work. If for example I square both sides of an equation, and then try to go back (using square root) that would give me the absolute values of both sides, which is not what I started with (unless I know that they are positive for sure), so I can conclude that some of the solutions I find might not work
This shows why you start solving such problems by trying out a few simple numbers. Right hand side has a cube root of x, so you try the first cube numbers: 0: nope, 1: nope, 8: bingo!
Given √{1+√(1+x)} = x^(1/3) Squaring both side 1+√(1+x)=x^(2/3) Rearranging the terms 1-x^(2/3) = √(1+x) Squaring both side 1 + x^(4/3) - 2x^(2/3) = 1+x Rearranging and cancelling terms x^(4/3) -2x^(2/3) - x^(3/3) = 0 Taking ∛x as 'm' so ⇒ m⁴ - 2m² - m³ = 0 Taking m² common m²(m²-m-2) = 0 m²(m-2)(m+1) = 0 So m=0 or -1 or 2 ⇒ ∛x = 0 or -1 or 2 ⇒x= 0 or -1 or 8 Putting x=0 ⇒1=0 which is not possible Putting x = -1 ⇒1 = ∛(-1) Which is not possible Putting x=8 ⇒2=2 *So x=2 is the only solution* But I have a doubt, if x=2 is the only solution then why we got 3 solution in first place ??
TLDR: brute-forcing this problem is not that hard, but this video's solution looks much cooler than mine. Here's how I've done it, by cubing both sides once and squaring both sides twice in total: you can see how I exploited the 'x must be real' condition to its limits. First I cubed and squared both sides. The left-hand side becomes (1 + sqrt(1 + x))³, the right-hand side becomes x². I unpacked the left-hand side and transferred all elements without the sqrt(1+x) term to the right-hand side. The resulting equation is: (4+x)*sqrt(1+x) = x² - 3x - 4. I modified the right-hand side into (x-4)(x+1). *x should be real number* so x >= -1, but when x=-1 the equation does not work. So x > -1. Square both sides, divide both sides with (x+1), and when I transferred all terms to one side I ended up with x³ - 8x = 0, or *x²(x-8) = 0.* x is 8.
I have two questions... How do we know that that this is in fact a quartic (p=4)? And I got the same extraneous solutions as well. Where do the other three extraneous solutions come from? (x=-1 and x=0, twice) Edit: Fwiw I raised to the 6th first, did some algebra, and then squared to get the quartic. The quartic was super easy since the x^1 and x^0 terms were 0.
x = 8 is the solution to the original equation, √(1 + √(1 + x)) = ∛x x = -1 is the solution to the equation -√(1 + √(1 + x)) = ∛x x = 0 is the solution to the equation √(1 - √(1 + x)) = ∛x Now consider the following procedure: _Square both sides of the equation, subtract 1 from both sides, then square both sides again._ If we apply this procedure to each of the three equations above, we get each time _the very same_ quartic equation (quartic in variable u = ∛x ): 1 + x = ((∛x)² - 1)² or 1 + u³ = (u² - 1)² which is more easily recognized as a quartic equation. That means that the solutions of this quartic equation includes the solutions of those three different equations, since each of those three different equations can be transformed into this quartic equation. (Your procedure raises both sides to the power of 6, then rearranges terms and squares both sides to get pretty much the same quartic. That's the same thing, I'm just using my version of the procedure because it's shorter and clearer for understanding. Well, _almost_ the same thing, since your approach leads to a quartic equation in x while my approach leads to a quartic equation in u = ∛x .) I hope that helps.
@@yurenchu Yes that helps, thank you! Basically I "lost" the ± sign on the square roots when I square them twice. Explicitly, our algebra works identically on these four equations: ±√(1+ ±√(1+x)) = ∛x. Cheers!
@@lreactor You're welcome! And you're exactly right! I just noticed that I made a little mistake in my previous reply; I should have written: x = 0 is the solution to the equations √(1 - √(1 + x)) = ∛x _and_ -√(1 - √(1 + x)) = ∛x (So there are indeed _four_ distinct equations that are all transformed into the same quartic equation by the same procedure.)
Let's elevate each side to the 6th power: ( 1 + √(1 + x) )^3 = x² Setting t=√(1 + x), we get the system: { x² = (1+t)^3 { t² = x+1 => t²-1 = x => (t-1)²(t+1)² = x² = (1+t)^3 => (t+1)² (t+1+(t-1)²) = 0 => (t+1)².t.(t-3) = 0 t = 3 is the only valid solution => x = 8
That would actually be to the power of 6. For the left side, it’s “first square, then cube”, and for the right side it’s “first cube, then square”. Remember the multiplication rule for power to a power.
I did the following: Raise to the power of 6 (since the LCM of 2 and 3 is 6) on both sides: (1 + √(1+x))³ = x² => 1 +3(1 + x) + 3 √(1 + x) + (1+x)√(1+x) = x² => (4 + 3x) + (x + 4)√(1 + x)=x² => (x + 4)√(1 + x) = x²-3x-4 => (x + 4)√(1 + x) = (x+1)(x-4) or x + 1 = 0 => x = - 1 => x + 4 = (x-4)√(1 + x) => (x+4)² = (x-4)²(1+x) => (x² - 8x + 16) (1+x) - (x² + 8x + 16) = 0 => x² - 8x + 16 + x³ - 8x² + 16x - x² - 8x - 16 = 0 => x³ - 8x² = 0 => x = 0 or x = 8 Check x = - 1 is not a solution, since the primary square root of 1 = 1, not -1. Check x = 0 is not a solution, since the right-hand side is 0 in the original equation, while the left is √2. So the only solution is x = 8.
I did u = sqrt(1 + x), x = u² - 1 And then v = sqrt(1 + u), u = v² - 1 Was able to do it in my head with no mistakes until I goofed at the end, thinking my check for x = -1 worked. I liked the x = u³ method presented in many of the comments better though. (Without the cubing stuff in the video). In retrospect, I think there could be some argument for x = -1, as we aren't dealing specifically with functions. Although I do recognize and accept the convention that the radical sign implies the positive root only. It's just weird that in the above context we accept negative values for cube roots, but not square roots.
It's not weird. We don't accept negative values (as outcome) for cube roots, _unless the argument of the cube root is also negative_ . But nobody says "cube root of 8 is -2 ", because that's simply incorrect. By the way, if you want to support the solution x = -1, then you should also support the solution x = 0 .
@@go_gazelle With x = 0 , the righthandside becomes ∛0 = 0 , and the lefthandside becomes √(1 + √(1 + 0)) = = √(1 + √(1)) ... and according to your logic in which x=-1 is supposedly a valid solution, √(1) = -1 would be a valid statement, hence ... = √(1 + (-1)) = √(0) = 0 which would match the righthandside.
@@yurenchu don't know why I didn't see that!! How about this argument for why neither incorrect solution is valid. Check x = -1: [-1, 1] ≠ -1 Check x = 0: [-√2, 0, √2] ≠ 0 Oh shoot. Then the actual solution wouldn't work either. Check x = 8: [-2, 2, -√(2)i, √(2)i] ≠ 2 This might be at least part of the reason why we don't include negative roots when simplifying radical expressions.
@@go_gazelle Nice analysis, LOL. However, if you want to consider complex-valued evaluations for the lefthandside with x=8, then please note that complex-valued evaluations also "exist" for the righthandside with x=-1 and with x=8 .
first raise both LHS and RHS to 6th power and then substitute 1+x = t^2 and then work out... it will be easy and 8 will com out as answer very easily no much tedious calculation.
I got the same 3 solutions right (in a different way: squaring the roots, and then I made y=3✓x; I solved for y, and then found the 3 values for x) But I didn't check if that was right substituting in the original equation😔
X=-1 works if you take the negative square root when you calculate sqrt(1). X=0 also works if you take the negative square root when you calculate sqrt(1-0).
@@macloko "A solution to the square root" is a nonsensical expression, a _non-sequitur_ , a "syntax error". We have square roots, and we have solutions to equations. But we don't have "a solution to the square root". An equation like z² = 3 has two solutions: they are z = +√3 and z = -√3 . z = +√3 represents one value (somewhere around 1.73205... ), and z = -√3 represents one, different, value (somewhere around -1.73205... ). If someone (or some math task) demanded you to quantify the positive solution to the equation (z-1)² = 3 , how would you do that concisely, in exact mathematical terms?
@@yurenchu Your point is well made. I should have said when we perform the square root operation, if we continue to evaluate the equation using the negative root as opposed to the positive root, the equation evaluates to be equal on both sides.
@@macloko Yes, that's true. However, the other person's reply still applies: You are _not_ allowed to do that, because the square root operator √x has been defined to refer to one particular solution of the equation z² = x (namely the positive solution, in case x is a positive real number), and the other solution is then described by -√x . The square root operator could have been defined the other way around ( √x being the negative solution, and -√x being the positive solution), it would have worked out just as correctly albeit somewhat clumsily, as long as √x is consistently defined and used as only one of the two solutions of the equation z² = x , never both solutions at the same time. But that would be a deviation from convention, and I see no advantage to it.
at first glance, I immediately squared both sides to get 1+sqrt(1+x) = x^(2/3) moving the +1 from LHS to RHS, sqrt(1+x) = x^(2/3) - 1 squaring both sides again, 1 + x = x^(4/3) - 2*1*x^(2/3) + 1 since there is +1 on both sides, they can cancel out so we get, x = x^(4/3) - 2*x^(2/3) moving the x from LHS to RHS and switching the sides(since we can do that) x^(4/3) - 2*x^(2/3) - x = 0 x^{1+(1/3)} - 2*x^{1-(1/3)} - x = 0 using exponent rules, it implies that, x*x^(1/3) -2*x*x^(-1/3) - x = 0 x{x^(1/3) - 2*x^(-1/3) - 1} = 0 so, x=0 seems like a solution here, but the second part of the equation makes absolutely no sense with x=0 and as seen in the video, it does not work. then, we say the second multiple of the LHS is equal to zero for the second case just for ease of understanding, I'll suppose t=x^(1/3) then the second multiple becomes: t - 2/t - 1 = 0 since t is not equal to 0, I can multiply t on both sides t^2 - 2 - t = 0 t^2 - t - 2= 0 [rearranging] t^2 - 2t + t - 2 = 0 t(t-2) + 1(t-2) = 0 (t+1)(t-2)=0 then, for the second case(since we had a case of x=0), we get t+1=0 then t=-1 x^1/3 = -1 I do not need to put it back in to know that it will not work. That is because x^1/3 is the RHS of the original equation and it CANNOT be negative since there is a square root on the LHS which prevents it from being negative. so the final case is t-2=0 t=2 x^(1/3)=2 x=8[by cubing both sides] therefore, x=8
Solution: x = 8 Derivation: √(1 + √(1 + x)) = ∛x ... let u = ∛x ... √(1 + √(1 + u³)) = u ... square both sides ... 1 + √(1 + u³) = u² √(1 + u³) = u² - 1 ... square both sides ... 1 + u³ = (u² - 1)² 1 + u³ = u⁴ - 2u² + 1 u³ = u⁴ - 2u² 0 = u⁴ - u³ - 2u² 0 = u²⋅(u² - u - 2) 0 = u²⋅(u+1)⋅(u-2) u² = 0 OR (u+1) = 0 OR (u-2) = 0 u = 0 OR u = -1 OR u = 2 ∛x = 0 OR ∛x = -1 OR ∛x = 2 x = 0³ = 0 OR x = (-1)³ = -1 OR x = 2³ = 8 x = 0 is not a solution to the original equation. (lefthandside would become √(1 + √(1 + 0)) = √(1 + √1) = √(1 + 1) = √2 , and righthandside would become ∛0 = 0 , which is not a match.) x = -1 is not a solution to the original equation. (lefthandside would become √(1 + √(1 + (-1))) = √(1 + √0) = √(1+0) = √1 = 1 , while righthandside would become ∛(-1) = -1 , which is not a match.) x = 8 is a solution to the original equation. (lefthandside becomes √(1 + √(1 + 8)) = √(1 + √9) = √(1 + 3) = √4 = 2 , and righthandside becomes ∛8 = 2 , which is a match.) Q.E.D.
I wonder ff it took me 1 minute to basically guess that x=8 just by trying it out with pen and paper (and I'm as bad at math as it gets) then I'm not sure if it's just luck or can you really just guess...?
Let Z=SQRT(1+X) So X=Z^2-1 Rewrite equation SQRT(1+Z) = CUBERT(Z^2-1) Cube both sides SQRT(1+Z) (SQRT(1+Z) (SQRT(1+Z) = Z^2-1 Simplify (Z+1) (SQRT(Z+1) = (Z+1) (Z-1) Divide both sides by (Z+1) SQRT(Z+1) = Z-1 Square both sides Z+1 = Z^2 - 2Z +1 Simplify Z^2 - 3Z = 0 Solutions for Z are Z=3 and Z=0 Since X=Z^2-1 Solutions for X are X=8 X= -1
x = 8 is the solution to the original equation, √(1 + √(1 + x)) = ∛x x = 0 is the solution to the equations √(1 - √(1 + x)) = ∛x and -√(1 - √(1 + x)) = ∛x We solved the original equation by applying the following procedure: _Square both sides of the equation, subtract 1 from both sides, then square both sides again._ However, when we apply this procedure to either of the three equations above, we get _the very same_ equation, namely: 1 + x = ((∛x)² - 1)² or if we substitute u = ∛x , 1 + u³ = (u² - 1)² which is more easily recognized as a quartic equation (quartic in variable u). That means that the solutions of this quartic equation includes the solutions of all three previous equations, since each of those three different equations can be transformed into this quartic equation. (In fact, the quartic equation also includes the solution of the equation -√(1 + √(1 + x)) = ∛x , which has the solution x = -1 . Note that this equation too is transformed into the same quartic equation, by the same procedure.) I hope that helps.
No need for difference of squares or sum of cubes. You don't need all of this. Just substitute like he did. Square like he did. Move square root to one side and everything else (u^2 - 1) to the other side. Square again. Move everything to one side. Take u^2 in front of the parenthesis and you'll get: u^2 (u^2 - u - 2) = 0. Solve the quadratic equation and you'll get u = -1 or u = 0 or u = 2. The rest is the same.
how would you get u^2 common from (u^3+1)-(u^2-1)^2=0 ?you wish to solve a fractional equation?it's not possible before taking out u+1 out then the constant terms cancels out in the other factor..then you finally know that u^2 can be taken out not before that
I don't know why all solutions presented are so ridiculously complicated. All one has to do is square both sides, rearrange, square both side again, cancel out the 1 on both sides, divide by x, substitute y=x^(1/3) , solve the quadratic and cube the root to find x.
Wow, thanks for mentioning me Presh. Thanks for solving this problem so beautifully. Kudos to you!
Glad to see ya here prof.
I think there's a simpler solution.
Let: x = y³
Square both sides: 1 + √(1+y³) = y²
Minus 1 and square: 1 + y³ = y⁴ - 2y² + 1
Rearrange and factor: 0 = y⁴ - y³ - 2y²
Factorise: 0 = y²(y² - y - 2)
Factorise: 0 = y²(y - 2)(y + 1)
Substitute: x = 0, x = 8, x = -1
Check: x = 8 is the only solution.
hey I did this in the same way too
technically all of them work if you consider sqrt(1)=+ -1
@@abhijiths5237 There'd be other solutions too if we considered 1+1=3.
However, since 1+1≠3 and √1≠-1, those solutions don't work.
@@beng4186 I'm talking about the multivariable roots. Let √x=1. So x²=1 and x can be +1 or -1 as -1² also equals 1. We consider square roots generally as +ve so most of the time we ignore -1. Haven't you heard of the cube roots of unity?
interestingly its only one solution, x=8, and no complex once
for fun tested √(1+√(1±x) )= (±x)¹/ ³
..the solutions is 8, -8, 1, -1 ...
∛is cubroot symbol (the YT font does a poor job displaying it clearly), using ¹/ ³ instead
I used a substitution for √(1 + x). This makes x = m^2 - 1. Sub in and raise both sides to the 6th power. Since m + 1 =\= 0, divide both sides by (m + 1)^2. Simple after that. -1 and 8, with -1 extraneous
Yup, that's exactly the way that I did it.
@@philstubblefield Same here. Instant simplification. No more nested root. Kinda obvious.
same here
Same
It's way simpler without utilizing the difference of squares or sum of cubes. The term 1 on both sides just cancels out, then you can factorize the polynomial and get smth like u^2(u^2-u-2). The quadratic is easily factorizable by guess to (u-2)(u+1).
Same for me. I solved it and then watched the video wondering the whole time WHY was he making it so complicated?!?
(Probably to throw in the formulas as a bonus lesson, I guess...)
The reason you cannot do what you suggest is that your solution requires you to divide both sides by u + 1. There is no guarantee that u -1, which results in dividing by zero.
@@salkabalani1482 no. If you just move everything to one side, you can factor without any division. And even if there was a division, we would just have to check u=1 separately.
Instead of any substitution, I did it by raising LHS and RHS to the power of 6 (LCM of 2 and 3), then ending with an equation containing only √(x+1) and other x terms, which then can be squared to get a polynomial equation, and solve them... It takes some steps to solve, but hey, its a solution...
If someone aa curious as me, there are always some ways to make extraneous solutions work.
x=-1 is solution if we assume external square root as a branch of multivalued function fixed by sqrt(1) = -1
x=0 is solution if internal square root is a branch fixed by the same thing as above.
So cool
x = t²+2t
LHS = √(1+t+1) = √(t+2)
RHS = ³√[t(t+2)]
Raising both sides to the 6th power,
(t+2)³=t²(t+2)² => (t+2)²(t²-t-2) = 0 and hence t = -2, -1 or 2 => x = 0, -1 or 8
A simple check shows x=8 is the only solution.
I loved the geometric representation at the beginning of the explanation. Makes it so much easier to visualize where these algebraic equations originated.
u=1+sqrt(1+x) is a way nicer substitution, for anyone who wants to try
Incredible. Another great one.
I had to pause the video at points but I understood the method. I'm glad you mentioned not to stop since after that long calculation I would have chosen all three numbers to be the final answer which is wrong.
It's easier than that: squaring both sides you get (1 + x)^(1/2) + 1 = x^(2/3). Since the product of (1 + x)^(1/2) + 1 and (1 + x)^(1/2) - 1 is x, we have that (1 + x)^(1/2) - 1 = x^(1/3). Subtracting the two equations gives x^(2/3) - x^(1/3) = 2. Letting y = x^(1/3) we get that y^2 - y - 2 = 0, which has solutions y = 2 and y = -1, corresponding to x = 8 and x = -1 respectively. Only the first fits the original equation, so x = 8 is the only solution.
"Since the product of (1 + x)^(1/2) + 1 and (1 + x)^(1/2) - 1 is x, we have that (1 + x)^(1/2) - 1 = x^(1/3)"
How did you get to that from that product?
@@maxhagenauer24 We are using that if ab = x and a = x^(2/3), then b = x/a = x/(x^(2/3)) = x^(1/3).
Where do you get the equation x^(2/3) - x^(1/2) = 2?
@@maxhagenauer24fooled me for a moment too:
Since the first equation gave us
1+sqrt(1+x) = x^2/3
[1+sqrt(1+x)] * x^1/3 = x
Now noting that
[sqrt(1+x)+1] [sqrt(1+x)-1] = x
Setting them equal we can divide by the like term and get
x^1/3 = sqrt(1+x) -1
@@ericanibal6222 Subtract the equation (1 + x)^(1/2) - 1 = x^(1/3) from the equation (1 + x)^(1/2) + 1 = x^(2/3)
By inspection X=8
I did that too, but it didn't clearly mean it was the only solution without a little analysis -- I looked at neighborhoods of the solution to convince myself, it was the only one.
Lmao I did this too!
I didn't even have to use substitution
After squaring both sides I got:
1+√(1+x)=³√x²
√(1+x)=³√x² -1
Squaring both sides again
1+x=(³√x² -1)²
1+x=³√x⁴ -2•³√x² +1
x=³√x⁴ -2•³√x²
Dividing both sides by ³√x²
x^(1-2/3)=³√x² -2
³√x=³√x² -2
Rearranging the terms to resemble a quadratic
(³√x)² -³√x -2=0
Factoring
(³√x -2)(³√x +1)=0
And I get
³√x=2 and ³√x=-1
So x=8,-1, but after checking them the only viable solution is x=8
1st! Really love your videos! They always help me to learn something new. Thanks ❤
To eliminate both square and cube roots, take the 6th power of both sides of the equation. Expanding and rearranging terms gives: (4+x)√(1+x) = x²-3x-4 = (x+1)(x-4). Dividing both
sides by √(1+x) (since x≠-1) gives 4+x = (x-4)√(1+x). Squaring again, expanding and collecting like terms gives: x³-8x² = 0, which has the obvious solution x=8. (The extraneous solutions
x=0 and x=-1 are solutions to similar equations with √ replaced by -√.)
Nice! But i got an even simpler solution, in which we dont have to deal with sum and difference of cubes, which inadvertently leads us to making a calculation error somewhere most of the time.
If we square the original equation (we will get x^2/3, but we don't have to worry about those weird exponents), take the 1 on the other side, and square it again to remove the root inside the root, then we get an equation in x with fractional powers
(the equation is : x^4/3 - 2x^2/3 - x) The equation sure looks scary but it can be simplified quite nicely by taking a substitution of cuberoot (x) = t.
After the substitution, we can factor in and solve the equation hence obtained extremely easily, and get t=0,0,2,-1 which subsequently gives us x = 0,0,8,-1 and checking back we can eliminate the redundant solutions and get x = 8 as a solution.
EASY RIGHT?
I did the calculation in the old-fashioned way, entirely without substitution. I raised both sides to the sixth power. (You can also first square both sides and then cube them both, if you prefer that.)
Then, I isolated the term with the root and squared the equation again.
One might assume that this results in a super unwieldy equation but after simplifying the terms, one is left with this:
7x³ + 8x² = x⁴
We can further simplify by dividing by x² (and testing whether x²=0 and therefore x=0 is a solution; it is of course not one), and we get this simple quadratic equation:
x² - 7x - 8 = 0
This equation has two solutions, x=-1 and x=8; testing shows that x=-1 is an extraneous solution but x=8 is valid.
And that's it! 😀
@kintagrama What I meant is the following:
We start with the equation as seen:
√[1+√(1+x)] = ∛x
Then, raising both sides to the 6th power (because we need to eliminate the square root on the left side and the cube root on the right side) gives us:
[1+√(1+x)]³ = x²
Expanding the cubed sum on the left hand side results in:
1³ + 3⋅1²⋅√(1+x) + 3⋅1⋅√(1+x)² + √(1+x)³ = x²
And now comes the isolation I was talking about. We got rid of the root on the right side in our very first step but there are still roots in the second and fourth addend. (The root in the third addend is canceled by the square.)
So we can rearrange the equation in a way that we have all addends with a root on one side and all others on the other side:
3 √(1+x) + √(1+x)³ = x² - 3 (1+x) - 1
Next, we can rewrite √(1+x)³ as √(1+x)² ⋅ √(1+x)¹ - and do a little clean-up on the right side. The root and square of the first factor again cancel out, leaving us with the same root we already have in the first addend:
3 √(1+x) + (1+x) √(1+x) = x² - 3x - 4
In a final step, we can now factor out the root on the left side:
(x+4) √(1+x) = x² - 3x - 4
And that's it!
This is what I meant by isolating the term with the root. The idea is that we can now square both sides without this in resulting in another root anywhere - something you can't do as long as you have a root in a sum. (For example, if you have 4+√x on one side, you can square this all day long but you'll always get another root: (4+√x)² = 4²+8√x+x. The first and third summand aren't roots but the middle one is. If you then group the root and non-root terms together (x+16)+(8√x) and square another, you'll again get two addends without roots but the middle one keeps the root. This will never change unless you can move all non-root addends on the other side.)
Does all of this make sense?
Just from the expected shapes of the functions I would only expect one solution if there is any. You have a cube root and something that will generally act like a fourth root. Those two only intersect at 0 and 1, but the adding of 1s will shift it around so I don’t expect them to meet up at both of those two spots anymore.
Not a solution, but the quick analysis so I can see if what I get in the end makes sense.
wow! i really like that visual explanation for the formulas, it really helped me understand!
I just saw x = 8 by inspection and I did some calculus reasoning to find that there probably aren't any more solutions
Me too!
Do the u substitution and rearrange till you get -u^4 + u^3 + 2u^2 = 0, factor out u^2 and solve for u to be 0, 2, and -1. Sub these back in to find x and you get that the only one that works is 8.
I raised both to the sixth power, then eventually got to:
(x+4)(x+1)^0.5 = x^2-3x-4
Squared both sides of that and got:
x^4-7x^3-8x^2=0
This factors to (x-8)(x+1)(x^2)=0. The rest follows the same way as 8:55 on.
It is possible to dispense with the extraneous solutions at the very start by noticing that the left-hand side of the original equation can't have a value of less than 1 for every possible value of x in the domain of that formula. This means that cbrt(x) can't be less than 1; that is, x can't be less than 1. Since x=8 is the only solution that fits this criterion, it's the only solution of the equation.
Does an extraneous solution mean an impossible solution?
extraneous solution means solutions obtained when we square or cube both sides but those solutions aren't actually true.
@asr2009 has the right of it. In general, extraneous solutions are introduced by the math used to solve the problem. For radical equations, extraneous solutions can occur when you raise the equation to an even power (this is because doing so destroys information about signs).
For a different type of extraneous solution, rational equations (where there is a variable in the denominator) can have extraneous solutions introduced by multiplying. When you plug those in to check, they generally cause division by zero.
In all cases, extraneous solutions are solutions that do not work in the original equation, but only in the transformed one.
@@davidtripp2118thanks!!
@@davidtripp2118 When I looked up the graphs of *√(1+√(1+x)) = x⅓* and it's transformed equation *x⁴ - 7x³ - 8x² = 0* on Desmos, It showed X = 8 for first equation and X = -1 & 8 for the second equation.
We can clearly see that X = 0 is also a solution for the second equation, so why does it eliminate that possibility?
The extra solutions only "work" if you take the negative square roots of the original equation -- not something done in formal mathematics!
For example, for x = 0 you can take the negative square root of the inner sqrt(1 + x) term to get -1, then add that to the 1 and get sqrt(0) = 0^1/3 = 0.
Also for x = -1, the inner sqrt term is zero (1 + -1), resulting in sqrt(1) = (-1)^1/3. Take the negative sqrt of 1 there to get -1 = -1.
It's issues like these that caused mathematicians to regulate using only the "principle square root" in equations, which is why the 0 and -1 solutions don't work in practice.
I did this with way less steps and not needing difference of cubes or squares. I just didn't do a u substitution and worked on getting rid of the squares, followed by some simplification and factoring. I was able to get the same answers in about about 9 lines of work. I do like that you can take advantage of the difference of cubes and squares if you want.
Thank goodness for UA-cam's speed control, as you go at rocket pace much too fast for me to follow along. Playing at 0.75x makes it actually possible to follow. If I miss one single step I'm screwed, that's why going so fast is a problem.
I wish you had more videos on stereometry, Presh! It's very enlightening.
Answer 3
Explanation:
Square both sides first
Then
Solve using substitute √1+x = t then substitute x= t^2 - 1 and make equation in 't' then solve for t quadratic equation then t=2 and x will equal to' 3'
So 3 is the answer
Love this solution and graphical representation of those formulae ❤
A cool use for the difference and sum of cubes formulas in my opinion is rationalising the denominator when it has a cube root
This was Finnish math olympiad problem Dr PK Math posted about a month ago.
Why not get to x^(4/3)-2x^(2/3)= x and substitute x=y^3. y^4-2y^2=y^3. Check for y=0 and then solve y^2-y=2. y=2 or -1 and x=8 or -1 but x=-1 is invalid.
yeah thats what i did lol
The answer is 8, simply done by trying trying out numbers that gave an integer answer on the left. So sqrt(x+1) must be integer. X=0 doesn't work, as we get sqrt(2) on the left, but the next candidate x=3 gives 2 on the left and the right. Of course, if the answer wasn't integer, this wouldn't work, but if time was critical in a test, worth a few seconds to try it.
I solved it but I think I made it much harder for myself. I didn't substitute at all and only brute forced it by cubing and squaring the entire equation and distributing the terms whenever possible. At one point I had to distribute two terms both with 3 different terms and thus ended up with 9 terms. I was surprised that in the end I was left with a quadratic equation that had a really simple discriminant. And after checking the solutions, those being -1 and 8, 8 turned out to be the correct solution. I'd like to know if anyone else went about the problem this way or if it's just a needlessly complicated process
Easiest way is to look 1- (1+x)=1 .which we can use to first square the exp and than take reciprocal.
1. Square -> 1+ sqrt(1+x) = x^2/3
2 .take reciprocal->
1-sqrt(1-x)=x^(1/3)=sqrt(1+sqrt(1+x))
3. Now square and solve
U get quadratic
I absolutely loved the visual explanation of the diff of squares and cubes! While I agree with many of the people here about the simpler way of doing it. I did the same all the while wondering if I was losing any mathematical rigor in the process (I have learned that you can’t cancel things willy nilly even tho’ it may be obvious). am curious to know why you did not go that route.
That is the most roundabout and incoherent solution video I've ever seen. This is what I did:
Square both sides and rearrange to get: sqrt(1+x) = x^(2/3) - 1
Square both sides again to get: 1+x = x^(4/3) - 2x^(2/3) + 1
Simplification yields: x^(4/3) - 2x^(2/3) - x = 0
Now make the following substitution: u = x^1/3
u^4 - u^3 - 2u^2 = 0 It's obvious that you can factor out u^2 here.
u^2 (u^2 - u - 2) = 0 u=0 is a root.
Solving the quadratic in parenthesis yields: u = (-1,2)
Now back-substitute to convert u's to x's. x = u^3
x = (0^3, (-1)^3, 2^3) = (0,-1,8)
Check the above values for x against the original problem equation to determine actual solution(s) and discard any extraneous roots.
When running that check, only one solution emerges: x = 8. Done.
square both sides, subtract 1 then square both sides again. set u = x^1/3 and it's an easy solution
I did this doing two substitutions, x = a^2-1, then a = b^2-1, then cubed both sides and solved to get b = -1,0,2 going back through the substitution chain, I get x = -1, 0, 8. -1 and 0 are extraneous, leaving 8.
To everyone complaining that he didn't use the most direct methodology to find the solution(s), you need to realize he's also teaching application of the sum and difference of squares in an easy problem so that people will know and remember to use it in more complex problems.
This is a teaching channel. Don't forget that.
First, let’s set ( y = \sqrt[3]{x} ). This means ( y^3 = x ). Substituting ( y ) into the equation, we get:
[ \sqrt[3]{1 + y} = y ]
Cubing both sides to eliminate the cube root, we have:
[ 1 + y = y^3 ]
Rearranging the equation, we get:
[ y^3 - y - 1 = 0 ]
Now, we need to solve this cubic equation for ( y ). Using numerical methods or algebraic manipulation, we find that one of the roots of this equation is ( y = 2 ).
Substituting ( y = 2 ) back into ( y^3 = x ), we get:
[ 2^3 = x ]
So, ( x = 8 ).
Therefore, the correct solution to the equation is ( x = 8 ).
I substituted x for a^3 and it worked really easy
Hey, please read this, I got this problem stuck in my head for the past month. If I draw a rectangle and on the long side of it I draw a square with the same dimensions as the long side and then I lengthen the side to infinity, the area of the square is infinity but also the one of the rectangle but the squares infinity is bigger than the infinity of the rectangle, why?
As others have pointed out, this solution is not the simplest. Here's what I did:
(1 + (1 + x)^(1/2))^(1/2) = x^(1/3)
1 + (1 + x)^(1/2) = x^(2/3) -Square both sides
(1 + x)^(1/2) = x^(2/3) - 1 -Move 1 to the other side
1 + x = x^(4/3) - 2x^(2/3) + 1 -Square both sides
x^(4/3) - x - 2x^(2/3) = 0 -Move everything to one side (1's cancel)
[x^(2/3)][x^(2/3) - x^(1/3) - 2] = 0 -Factor out x^(2/3)
[x^(2/3)][x^(1/3) - 2][x^(1/3) + 1] = 0 -Factor the quadratic in terms of x^(1/3)
Then each factor can be set equal to 0 and solved for x. The same solutions of 0, 8, and -1 emerge. Only 8 works.
I solved this in a simpler manner, I think.
I used the same u substitution and squared both sides, getting sqrt(1 - u^3) = u^2 - 1. I squared both sides again to get 1 + u^3 = u^4 - 2u^2 + 1. The ones cancel, and moving everything to the left we have u^4 - u^3 - 2u^2 = 0. This factors to (u^2)(u^2 - u - 2) = 0
Note that u cannot be negative, as the cube root of x is a principal square root. Also note that u cannot be 0 because we can test it in the original equation and show it will not work. Therefore u is strictly positive.
Since u cannot be zero, we can divide both sides by u^2 with no worries, leaving is with the quadratic which is easy and yields u = 2 or -1. u is strictly positive, so u = 2, and since x = u^3, x = 8.
Simple try to insert a couple of third power numbers give us 8 in 20 seconds or so, why so long way so solve?
X=8 at first glance 😂
But were you sure it was the only solution?
@@Kero-zc5tc just check cube numbers as in most of the case answer comes out to be the simplest one
Seriously!
Wolphram alpha, Mathway and similar apps can solve it very nicely.
@@himanshuuu6361 I think you missed my point, I was saying you don’t know if 8 is the only solution I was not at all referring to how to guess a solution
Well done bro
That is easy sum but thanks for explaining this sum for us ❤
Can someone explain to me if, and if so, how, we can make sure the value we got is true without the need of substitution?
Retrace your steps and do them in reverse, if there is no problem then all the solutions should work.
If for example I square both sides of an equation, and then try to go back (using square root) that would give me the absolute values of both sides, which is not what I started with (unless I know that they are positive for sure), so I can conclude that some of the solutions I find might not work
By Mihailescu's theorem, the only perfect powers that differ by 1 are 8 and 9. So there's only one candidate to check.
There's a simpler way to do this.
Let ∛x = u → x = u³
√(1 + √(1+x)) = ∛x
√(1 + √(1+u³)) = u
Square both sides:
1 + √(1+u³) = u²
√(1+u³) = u² − 1
Square both sides again:
1 + u³ = u⁴ − 2u² + 1
u⁴ − u³ − 2u² = 0
Now we are left with a quartic polynomial that's quite simple to factor:
u² (u² − u − 2) = 0
u² (u + 1) (u − 2) = 0
u = 0, −1, 2
x = u³ = 0, −1, 8
Now we check each solution to eliminate any extraneous solutions:
x = 0 → √(1 + √(1+x)) = √(1 + √(1+0)) = √(1+1) = √2
→ ∛x = ∛0 = 0 ╪ √2
x = −1 → √(1 + √(1+x)) = √(1 + √(1−1)) = √(1+0) = 1
→ ∛x = ∛−1 = −1 ╪ 1
x = 8 → √(1 + √(1+x)) = √(1 + √(1+8)) = √(1+3) = 2
→ ∛x = ∛8 = 2 ⇒ OK
The only solution is
*x = 8*
Love this stuff
This shows why you start solving such problems by trying out a few simple numbers. Right hand side has a cube root of x, so you try the first cube numbers: 0: nope, 1: nope, 8: bingo!
Well, that is only true if you assume there's going to be integer solutions. Of course it doesn't take long to try first few integers.
Square root of 1 can also be -1 so why isn't X = -1 a valid solution?
bait used to be believable
Very good video. Thanks
It literally took me 5 seconds to solve this in my head by inspection!
Given
√{1+√(1+x)} = x^(1/3)
Squaring both side
1+√(1+x)=x^(2/3)
Rearranging the terms
1-x^(2/3) = √(1+x)
Squaring both side
1 + x^(4/3) - 2x^(2/3) = 1+x
Rearranging and cancelling terms
x^(4/3) -2x^(2/3) - x^(3/3) = 0
Taking ∛x as 'm'
so ⇒ m⁴ - 2m² - m³ = 0
Taking m² common
m²(m²-m-2) = 0
m²(m-2)(m+1) = 0
So m=0 or -1 or 2
⇒ ∛x = 0 or -1 or 2
⇒x= 0 or -1 or 8
Putting x=0
⇒1=0 which is not possible
Putting x = -1
⇒1 = ∛(-1) Which is not possible
Putting x=8
⇒2=2
*So x=2 is the only solution*
But I have a doubt, if x=2 is the only solution then why we got 3 solution in first place ??
TLDR: brute-forcing this problem is not that hard, but this video's solution looks much cooler than mine.
Here's how I've done it, by cubing both sides once and squaring both sides twice in total: you can see how I exploited the 'x must be real' condition to its limits. First I cubed and squared both sides. The left-hand side becomes (1 + sqrt(1 + x))³, the right-hand side becomes x². I unpacked the left-hand side and transferred all elements without the sqrt(1+x) term to the right-hand side. The resulting equation is: (4+x)*sqrt(1+x) = x² - 3x - 4. I modified the right-hand side into (x-4)(x+1). *x should be real number* so x >= -1, but when x=-1 the equation does not work. So x > -1. Square both sides, divide both sides with (x+1), and when I transferred all terms to one side I ended up with x³ - 8x = 0, or *x²(x-8) = 0.*
x is 8.
I have two questions... How do we know that that this is in fact a quartic (p=4)? And I got the same extraneous solutions as well. Where do the other three extraneous solutions come from? (x=-1 and x=0, twice)
Edit: Fwiw I raised to the 6th first, did some algebra, and then squared to get the quartic. The quartic was super easy since the x^1 and x^0 terms were 0.
x = 8 is the solution to the original equation,
√(1 + √(1 + x)) = ∛x
x = -1 is the solution to the equation
-√(1 + √(1 + x)) = ∛x
x = 0 is the solution to the equation
√(1 - √(1 + x)) = ∛x
Now consider the following procedure: _Square both sides of the equation, subtract 1 from both sides, then square both sides again._
If we apply this procedure to each of the three equations above, we get each time _the very same_ quartic equation (quartic in variable u = ∛x ):
1 + x = ((∛x)² - 1)²
or
1 + u³ = (u² - 1)² which is more easily recognized as a quartic equation.
That means that the solutions of this quartic equation includes the solutions of those three different equations, since each of those three different equations can be transformed into this quartic equation.
(Your procedure raises both sides to the power of 6, then rearranges terms and squares both sides to get pretty much the same quartic. That's the same thing, I'm just using my version of the procedure because it's shorter and clearer for understanding. Well, _almost_ the same thing, since your approach leads to a quartic equation in x while my approach leads to a quartic equation in u = ∛x .)
I hope that helps.
@@yurenchu Yes that helps, thank you! Basically I "lost" the ± sign on the square roots when I square them twice. Explicitly, our algebra works identically on these four equations:
±√(1+ ±√(1+x)) = ∛x. Cheers!
@@lreactor You're welcome!
And you're exactly right! I just noticed that I made a little mistake in my previous reply; I should have written:
x = 0 is the solution to the equations
√(1 - √(1 + x)) = ∛x
_and_
-√(1 - √(1 + x)) = ∛x
(So there are indeed _four_ distinct equations that are all transformed into the same quartic equation by the same procedure.)
Let's elevate each side to the 6th power:
( 1 + √(1 + x) )^3 = x²
Setting t=√(1 + x), we get the system:
{ x² = (1+t)^3
{ t² = x+1
=> t²-1 = x
=> (t-1)²(t+1)² = x² = (1+t)^3
=> (t+1)² (t+1+(t-1)²) = 0
=> (t+1)².t.(t-3) = 0
t = 3 is the only valid solution
=> x = 8
That would actually be to the power of 6. For the left side, it’s “first square, then cube”, and for the right side it’s “first cube, then square”. Remember the multiplication rule for power to a power.
@@bobross7473 Ah you're right! My bad. I will correct that.
Beautiful.. thanks..
Can anyone explain why did we get extraneous solution like -1 and 0 even after solving completely correct
Please tackle this one 🙏🙏
if a= √3 +√4 +√5 find a⁴ - 8a³ +8a² +32a
Use chatgpt
Answer: 44
a = √3 + √4 + √5
a = √3 + 2 + √5
a - 2 = √3 + √5
(a - 2)² = (√3 + √5)²
a² - 4a + 4 = (3 + 2√15 + 5)
a² - 4a + 4 = 8 + 2√15
a² - 4a - 4 = 2√15
(a² - 4a - 4)² = (2√15)²
a⁴ + (-4a)² + (-4)² + 2*[(a²)*(-4a) + (a²)*(-4) + (-4a)*(-4)] = 60
a⁴ + 16a² + 16 + 2*[ (-4a³) + (-4a²) + (16a)] = 60
a⁴ + 16a² + 16 - 8a³ - 8a² + 32a = 60
a⁴ - 8a³ + 8a² + 32a + 16 = 60
a⁴ - 8a³ + 8a² + 32a = 44
By the way, the four solutions to the equation
a⁴ - 8a³ + 8a² + 32a = 44
are
a = √4 + √3 + √5
a = √4 - √3 - √5
a = √4 + √3 - √5
a = √4 - √3 + √5
@@yurenchu Good 👍👍 keep up the good work.
@@iMvJ27 LOL! 😊
Thanks for the encouragement, and for posting this interesting exercise in the first place!
I did the following:
Raise to the power of 6 (since the LCM of 2 and 3 is 6) on both sides:
(1 + √(1+x))³ = x²
=> 1 +3(1 + x) + 3 √(1 + x) + (1+x)√(1+x) = x²
=> (4 + 3x) + (x + 4)√(1 + x)=x²
=> (x + 4)√(1 + x) = x²-3x-4
=> (x + 4)√(1 + x) = (x+1)(x-4) or x + 1 = 0 => x = - 1
=> x + 4 = (x-4)√(1 + x)
=> (x+4)² = (x-4)²(1+x)
=> (x² - 8x + 16) (1+x) - (x² + 8x + 16) = 0
=> x² - 8x + 16 + x³ - 8x² + 16x - x² - 8x - 16 = 0
=> x³ - 8x² = 0
=> x = 0 or x = 8
Check x = - 1 is not a solution, since the primary square root of 1 = 1, not -1.
Check x = 0 is not a solution, since the right-hand side is 0 in the original equation, while the left is √2.
So the only solution is x = 8.
i tried the same but only made it to the second line and gave up. didnt think of moving 4+3x to the rhs
I did u = sqrt(1 + x), x = u² - 1
And then v = sqrt(1 + u), u = v² - 1
Was able to do it in my head with no mistakes until I goofed at the end, thinking my check for x = -1 worked.
I liked the x = u³ method presented in many of the comments better though. (Without the cubing stuff in the video).
In retrospect, I think there could be some argument for x = -1, as we aren't dealing specifically with functions. Although I do recognize and accept the convention that the radical sign implies the positive root only.
It's just weird that in the above context we accept negative values for cube roots, but not square roots.
It's not weird. We don't accept negative values (as outcome) for cube roots, _unless the argument of the cube root is also negative_ . But nobody says "cube root of 8 is -2 ", because that's simply incorrect.
By the way, if you want to support the solution x = -1, then you should also support the solution x = 0 .
@@yurenchu explain the argument (whether or not you like it) of how x = 0 works.
@@go_gazelle With x = 0 , the righthandside becomes ∛0 = 0 , and the lefthandside becomes
√(1 + √(1 + 0)) =
= √(1 + √(1))
... and according to your logic in which x=-1 is supposedly a valid solution, √(1) = -1 would be a valid statement, hence ...
= √(1 + (-1))
= √(0)
= 0
which would match the righthandside.
@@yurenchu don't know why I didn't see that!!
How about this argument for why neither incorrect solution is valid.
Check x = -1:
[-1, 1] ≠ -1
Check x = 0:
[-√2, 0, √2] ≠ 0
Oh shoot. Then the actual solution wouldn't work either.
Check x = 8:
[-2, 2, -√(2)i, √(2)i] ≠ 2
This might be at least part of the reason why we don't include negative roots when simplifying radical expressions.
@@go_gazelle Nice analysis, LOL. However, if you want to consider complex-valued evaluations for the lefthandside with x=8, then please note that complex-valued evaluations also "exist" for the righthandside with x=-1 and with x=8 .
try substituting -1 and 0 to the original equation and see if its true. (8 is true)
sqr(1) could be -1, then, the fist option could be a valid real solution.
first raise both LHS and RHS to 6th power and then substitute 1+x = t^2 and then work out... it will be easy and 8 will com out as answer very easily no much tedious calculation.
8
[ sqrt 1 + sqrt ( 1+x)]^6 = (cube root of 3)^6 raised both sides to the 6th power
[1 + sqrt ( 1 + x ]^3 = x^2
let sqrt (1 + x) = n
then 1 + x = n^2
and x = n^2 -1
Hence, (1 + n)^3 = (n^2 -1)^2
1 + n^3 + 3n + 3n^2= n^4 -2n^2 + 1
0 = n^4- n^3 -5n^2 - 3n
0 = n^3 - n^2 -5n -3 (divide both sides by n
0= n^3 -3n - n^2 - 2n -3
0 = n^3- 3n- (n-3)(n+1)
0 = (n+1) * (n-3)n+1)
n=-1 n=-1 and n=3
Hence, sqrt (1+ x) =3
1+ x =9
x=8 Answer
sqrt (1+ x) also = -1
1+ x =1
x=0 but 0 does not satisfy the equation.
Is it just me, or does everyone glance at the length of the video to estimate the difficulty of the problem? : )
Nope, not just you.
I got the same 3 solutions right (in a different way: squaring the roots, and then I made y=3✓x; I solved for y, and then found the 3 values for x)
But I didn't check if that was right substituting in the original equation😔
X=-1 works if you take the negative square root when you calculate sqrt(1). X=0 also works if you take the negative square root when you calculate sqrt(1-0).
However, you're *NOT* allowed to do that, so you should not bring that up.
Why not? It’s a solution to the square root?
@@macloko "A solution to the square root" is a nonsensical expression, a _non-sequitur_ , a "syntax error". We have square roots, and we have solutions to equations. But we don't have "a solution to the square root".
An equation like z² = 3 has two solutions: they are z = +√3 and z = -√3 .
z = +√3 represents one value (somewhere around 1.73205... ), and
z = -√3 represents one, different, value (somewhere around -1.73205... ).
If someone (or some math task) demanded you to quantify the positive solution to the equation (z-1)² = 3 , how would you do that concisely, in exact mathematical terms?
@@yurenchu Your point is well made. I should have said when we perform the square root operation, if we continue to evaluate the equation using the negative root as opposed to the positive root, the equation evaluates to be equal on both sides.
@@macloko Yes, that's true. However, the other person's reply still applies: You are _not_ allowed to do that, because the square root operator √x has been defined to refer to one particular solution of the equation z² = x (namely the positive solution, in case x is a positive real number), and the other solution is then described by -√x .
The square root operator could have been defined the other way around ( √x being the negative solution, and -√x being the positive solution), it would have worked out just as correctly albeit somewhat clumsily, as long as √x is consistently defined and used as only one of the two solutions of the equation z² = x , never both solutions at the same time. But that would be a deviation from convention, and I see no advantage to it.
at first glance, I immediately squared both sides to get 1+sqrt(1+x) = x^(2/3)
moving the +1 from LHS to RHS, sqrt(1+x) = x^(2/3) - 1
squaring both sides again, 1 + x = x^(4/3) - 2*1*x^(2/3) + 1
since there is +1 on both sides, they can cancel out
so we get, x = x^(4/3) - 2*x^(2/3)
moving the x from LHS to RHS and switching the sides(since we can do that)
x^(4/3) - 2*x^(2/3) - x = 0
x^{1+(1/3)} - 2*x^{1-(1/3)} - x = 0
using exponent rules, it implies that,
x*x^(1/3) -2*x*x^(-1/3) - x = 0
x{x^(1/3) - 2*x^(-1/3) - 1} = 0
so, x=0 seems like a solution here, but the second part of the equation makes absolutely no sense with x=0 and as seen in the video, it does not work.
then, we say the second multiple of the LHS is equal to zero for the second case
just for ease of understanding, I'll suppose t=x^(1/3)
then the second multiple becomes:
t - 2/t - 1 = 0
since t is not equal to 0, I can multiply t on both sides
t^2 - 2 - t = 0
t^2 - t - 2= 0 [rearranging]
t^2 - 2t + t - 2 = 0
t(t-2) + 1(t-2) = 0
(t+1)(t-2)=0
then, for the second case(since we had a case of x=0), we get t+1=0
then t=-1
x^1/3 = -1
I do not need to put it back in to know that it will not work. That is because x^1/3 is the RHS of the original equation and it CANNOT be negative since there is a square root on the LHS which prevents it from being negative.
so the final case is t-2=0
t=2
x^(1/3)=2
x=8[by cubing both sides]
therefore, x=8
Omg I graduated high school without knowing where the formula for the diff of square and cube came from! This came thirteen years too late for me. 😂
Fantastic
I did it in my head in about 20 - 30 seconds by substituting likley numbers.
Just trying the first few easy integers, the answer is 8. But I doubt that’s what was meant by “solve the equation”.
Why cant root of 1 be (-1)?
Very Simple question
Nice sum
Domain of the function was x>0 anyway, need not check for x = -1, 0
What function?
The domain of the lefthandside expression, √(1 + √(1+x)) , is x ≥ -1 (and not x>0) .
x=8
Solved it by guessing :D
Solution: x = 8
Derivation:
√(1 + √(1 + x)) = ∛x
... let u = ∛x ...
√(1 + √(1 + u³)) = u
... square both sides ...
1 + √(1 + u³) = u²
√(1 + u³) = u² - 1
... square both sides ...
1 + u³ = (u² - 1)²
1 + u³ = u⁴ - 2u² + 1
u³ = u⁴ - 2u²
0 = u⁴ - u³ - 2u²
0 = u²⋅(u² - u - 2)
0 = u²⋅(u+1)⋅(u-2)
u² = 0 OR (u+1) = 0 OR (u-2) = 0
u = 0 OR u = -1 OR u = 2
∛x = 0 OR ∛x = -1 OR ∛x = 2
x = 0³ = 0 OR x = (-1)³ = -1 OR x = 2³ = 8
x = 0 is not a solution to the original equation.
(lefthandside would become √(1 + √(1 + 0)) = √(1 + √1) = √(1 + 1) = √2 ,
and righthandside would become ∛0 = 0 , which is not a match.)
x = -1 is not a solution to the original equation.
(lefthandside would become √(1 + √(1 + (-1))) = √(1 + √0) = √(1+0) = √1 = 1 ,
while righthandside would become ∛(-1) = -1 , which is not a match.)
x = 8 is a solution to the original equation.
(lefthandside becomes √(1 + √(1 + 8)) = √(1 + √9) = √(1 + 3) = √4 = 2 ,
and righthandside becomes ∛8 = 2 , which is a match.)
Q.E.D.
Why are there wrong solutions for x?
Square both side to solve algebrically
Damn it was deep
Wow! I pretty much guessed that x could be 8, and I was right. I did none of that extra math.
X=8 my friend by inspection
I wonder ff it took me 1 minute to basically guess that x=8 just by trying it out with pen and paper (and I'm as bad at math as it gets) then I'm not sure if it's just luck or can you really just guess...?
Let Z=SQRT(1+X) So X=Z^2-1
Rewrite equation SQRT(1+Z) = CUBERT(Z^2-1)
Cube both sides SQRT(1+Z) (SQRT(1+Z) (SQRT(1+Z) = Z^2-1
Simplify (Z+1) (SQRT(Z+1) = (Z+1) (Z-1)
Divide both sides by (Z+1) SQRT(Z+1) = Z-1
Square both sides Z+1 = Z^2 - 2Z +1
Simplify Z^2 - 3Z = 0 Solutions for Z are Z=3 and Z=0
Since X=Z^2-1 Solutions for X are X=8 X= -1
Trial and error
My first guess was x=8😅
I graphed it and came up with 8.
For me is strange why solving the equation we found a solution inside the domain of the x (x=0) and suddenly this solution is not right...
x = 8 is the solution to the original equation,
√(1 + √(1 + x)) = ∛x
x = 0 is the solution to the equations
√(1 - √(1 + x)) = ∛x
and
-√(1 - √(1 + x)) = ∛x
We solved the original equation by applying the following procedure: _Square both sides of the equation, subtract 1 from both sides, then square both sides again._
However, when we apply this procedure to either of the three equations above, we get _the very same_ equation, namely:
1 + x = ((∛x)² - 1)²
or if we substitute u = ∛x ,
1 + u³ = (u² - 1)² which is more easily recognized as a quartic equation (quartic in variable u).
That means that the solutions of this quartic equation includes the solutions of all three previous equations, since each of those three different equations can be transformed into this quartic equation. (In fact, the quartic equation also includes the solution of the equation -√(1 + √(1 + x)) = ∛x , which has the solution x = -1 . Note that this equation too is transformed into the same quartic equation, by the same procedure.)
I hope that helps.
I solved by substituting root(x+1) with t and then after solving for t. I got x=8 as the answer directly😅.
No need for difference of squares or sum of cubes. You don't need all of this. Just substitute like he did. Square like he did. Move square root to one side and everything else (u^2 - 1) to the other side. Square again. Move everything to one side. Take u^2 in front of the parenthesis and you'll get: u^2 (u^2 - u - 2) = 0. Solve the quadratic equation and you'll get u = -1 or u = 0 or u = 2. The rest is the same.
how would you get u^2 common from (u^3+1)-(u^2-1)^2=0 ?you wish to solve a fractional equation?it's not possible before taking out u+1 out then the constant terms cancels out in the other factor..then you finally know that u^2 can be taken out not before that
@@mrityunjaykumar4202 Just calculate (u^2-1)^2 = u^4 - 2 u^2 + 1 and ones will cancel out. u^4 - u^3 - 2 u^2 = u^2 (u^2 - u - 2) = u^2 (u + 1) (u - 2)
I don't know why all solutions presented are so ridiculously complicated. All one has to do is square both sides, rearrange, square both side again, cancel out the 1 on both sides, divide by x, substitute y=x^(1/3) , solve the quadratic and cube the root to find x.
If you don't know why, then ask questions so that you'll understand.
@@corpsie666did you actually read my comment?
Hmm, improving. At first glance I thought, either 4 or 16. Not that far off.
yes. i can.
Why oh why use geometry???