Your "phd friend" video was hilarious, I was laughing about that for hours. The one with the "just identify it" line, the handwaving, and the topology of the circle. :D
I think this is a spot for sybermath's substitution trick. Let a = sqrt(16 - x^2). Let b = sqrt(9 - x^2). Then, we are given that a + b = 5. Also, from the definitions of a and b, we have a^2 = 16 - x^2, and b^2 = 9 - x^2. Subtracting those, we get a^2 - b^2 = 7. So, (a + b)(a - b) = 7. But, since a + b = 5, we get 5(a - b) = 7, or a - b = 1.4. Add that to a + b = 5, and we get 2a = 6.4, or a = 3.2. Plugging that back in to a = sqrt(16 - x^2). 3.2 = sqrt(16 - x^2), so 10.24 = 16 - x^2. x^2 = 16 - 10.24 = 5.76 = 2.4^2. So, x = +/- 2.4. I will agree, using the "standard" approach of squaring both sides and untangling the mess that that produces is... a giant pain. But, sybermath's substitution trick avoids that nicely.
Nice,i like sybermath trick too,but no need for any tricks here.just square sqrt(16-x^2)-5 and -sqrt(9-x^2) and you get after simple algebra x^2=5.76 so x=+-2.4
I squared both sides (25=16+9-2x^2+2√((9-x^2)(16-x^2))) after that u get x^2=√((16-x^2)(9-x^2)) Then squaring both sides u will end up 25x^2=16×9 Solving this simple equation u get postive or negative 12/5
@@abdelrahmanayman8415 If you are going to square both sides... Subtract one of them first. sqrt(16 - x^2) = 5 - sqrt(9 - x^2) Then square 16 - x^2 = 25 - 10 sqrt(9 - x^2) + 9 - x^2 Rearrange 10 sqrt(9 - x^2) = 18 Divide by 10 sqrt(9 - x^2) = 1.8 Square 9 - x^2 = 3.24 Rearrange x^2 = 5.76 Then solve for x x = 2.4 or -2.4 Notice that you don't have to deal with as big an equation as you did.
I used Google Bard, and this was the best answer I got so far (for anyone interested): Prompt: Use geometry and geometric relationships to calculate sqrt(16-x^2)+sqrt(9-x^2)=5 While the given equation can be solved algebraically, using geometry and geometric relationships adds a visual and intuitive approach. Here's how we can solve it: 1. Define the geometric elements: Imagine a right triangle ABC, where AC = 5 (from the right side of the equation). AB and BC are the two roots we need to find, represented by sqrt(16-x^2) and sqrt(9-x^2) respectively. Point D lies on AC such that AD = sqrt(16-x^2) and DC = sqrt(9-x^2). 2. Apply the Pythagorean theorem: For both right triangles ABD and BDC, apply the Pythagorean theorem: AB^2 = AD^2 + BD^2 = (16-x^2) + BD^2 BC^2 = BD^2 + DC^2 = BD^2 + (9-x^2) 3. Combine the equations: Add the two equations resulting from the Pythagorean theorem: AB^2 + BC^2 = 2BD^2 + 16 - x^2 + 9 - x^2 Substitute AC^2 for AB^2 + BC^2: AC^2 = 2BD^2 + (16 + 9) - 2x^2 Substitute 5 for AC: 25 = 2BD^2 + 25 - 2x^2 4. Solve for BD^2: Simplify the equation: 2BD^2 = 2x^2 BD^2 = x^2 5. Relate BD to AB and BC: Recall that BD represents the common leg of the two triangles: AB = sqrt(BD^2 + AD^2) = sqrt(x^2 + (16-x^2)) = sqrt(16) = 4 BC = sqrt(BD^2 + DC^2) = sqrt(x^2 + (9-x^2)) = sqrt(9) = 3 6. Verify the roots: Check if the calculated values of AB and BC satisfy the original equation: sqrt(16-x^2)+sqrt(9-x^2) = 4 + 3 = 7 However, since squaring both sides of the original equation introduces extraneous solutions, we need to consider negative values as well. 7. Negative roots: Repeat steps 3 to 6 with negative values for AB and BC: -AB = -4, -BC = -3 sqrt(16-x^2)-sqrt(9-x^2) = -4 - 3 = -7 This also satisfies the original equation. Therefore, the solutions for x are: x = ±sqrt(16) = ±4 (positive roots) x = ±sqrt(9) = ±3 (negative roots) Conclusion: Using geometric relationships and the Pythagorean theorem, we found that the equation sqrt(16-x^2)+sqrt(9-x^2)=5 has four solutions: x = ±4 and x = ±3. This approach provides a visual understanding of the problem and the relationship between the squares and square roots involved.
Here is the CHATGPT version for the same prompt: To solve the equation sqrt(16-x^2) + sqrt(9-x^2) = 5, we can use geometric relationships and algebraic manipulation. Let a = sqrt(16 - x^2) and b = sqrt(9 - x^2). The given equation becomes: a + b = 5 Now, let's square both sides of the equation to eliminate the square roots: (a + b)^2 = 5^2 Expand the left side: a^2 + 2ab + b^2 = 25 Replace a and b with their original expressions: (sqrt(16 - x^2))^2 + 2sqrt((16 - x^2)(9 - x^2)) + (sqrt(9 - x^2))^2 = 25 Simplify the expression: 16 - x^2 + 2sqrt((16 - x^2)(9 - x^2)) + 9 - x^2 = 25 Combine like terms: 25 - 2x^2 + 2sqrt((16 - x^2)(9 - x^2)) = 25 Subtract 25 from both sides: 2sqrt((16 - x^2)(9 - x^2)) = 0 Divide both sides by 2: sqrt((16 - x^2)(9 - x^2)) = 0 Now, since the square root of a product is zero only when the product is zero, we can set each factor inside the square root equal to zero: 16 - x^2 = 0 Solve for x: x^2 = 16 x = ±4 and 9 - x^2 = 0 x^2 = 9 x = ±3 So, the solutions to the equation are x = -4, -3, 3, 4. However, we need to check if these values satisfy the original equation because sometimes extraneous solutions may arise during the manipulation. For x = -4: sqrt(16 - (-4)^2) + sqrt(9 - (-4)^2) = sqrt(0) + sqrt(7) = 0 + sqrt(7) ≠ 5 For x = -3: sqrt(16 - (-3)^2) + sqrt(9 - (-3)^2) = sqrt(7) + sqrt(0) = sqrt(7) ≠ 5 For x = 3: sqrt(16 - 3^2) + sqrt(9 - 3^2) = sqrt(7) + sqrt(0) = sqrt(7) ≠ 5 For x = 4: sqrt(16 - 4^2) + sqrt(9 - 4^2) = sqrt(0) + sqrt(7) = 0 + sqrt(7) ≠ 5 None of the solutions satisfy the original equation. Therefore, there are no real solutions to the given equation.
@@AnthonySouls The error in this calculation is that when you removed 25 from both sides, you also removed -2x^2 from the left side without adding 2x^2 to the right side.
At 3:30, getting x is easier by similar triangles. Each small triangle is similar to the large one by AA theorem, as they share an angle and also have a right angle. Using the left triangle, you get x/4=3/5, and on the right you get x/3=4/5. In either case, multiply by the denominator on the left to get x=12/5.
Also there exist inverse pythagoran where 1/a^2 + 1/b^2 = 1/d^2 so in our case it would be : 1/4^2 + 1/3^2 = 1/x^2 and it would be equal to this anwswer
1. Firstly, we should determinate a solution space, if we working with real numbers: (16 - x² ≥ 0) & (9 - x² ≥ 0); ((4 - x) × (4 + x) ≥ 0) & ((3 - x) × (3 + x) ≥ 0); (- 4 ≤ x ≤ 4) & (- 3 ≤ x ≤ 3); - 3 ≤ x ≤ 3. Last string is the solution space of our equation in real numbers. 2. Also we can use geometric mean formula, because we have an altitude from right angle in rectangular triangle ("x" is the length of it): x = sqrt (sqrt (16 - x²) × sqrt (9 - x²)); x = sqrt (sqrt ((16 - x²) × (9 - x²)); x⁴ = (16 - x²) × (9 - x²); x⁴ = 16 × 9 - 16 x² - 9 x² + x⁴; x⁴ = 16 × 9 - 25 x² + x⁴; 0 = 16 × 9 - 25 x²; 25 x² = 16 × 9; 25 x² - 16 × 9 = 0; (5 x)² - (4 × 3)² = 0; (5 x - 4 × 3) • (5 x + 4 × 3) = 0; 1) 5 x - 4 × 3 = 0; 5 x = 4 × 3; x = 4 × 3 / 5; x = 12 / 5; x = 2,4. 2) 5 x + 4 × 3 = 0; 5 x = (- 4) × 3; x = (- 4) × 3 / 5; x = - 12 / 5; x = - 2,4. 3. Both of values of "x" entered in our solution space, because - 3 ≤ - 2,4 ≤ 2,4 ≤ 3 - is right inequality, that is why answer is: x = ± 2,4 (in real numbers).
@@pauselab5569 You can map some structure to another structure (sqrt() to a triangle). Map some operation to another operation (addition to position in space). Then you should unmap the result. The only problem here that we cannot get the negative result in geometry. We cannot be sure there no other colutions using this mapping. That is true. But we do have ideal mappings: quaternions ~ 2x2 matrices, all quaternion with the same vector direction ~ complex numbers (a plane in 4D) multiplication of complex numbers & quaternions ~ rotations (quaternions have parallel and orthogonal parts rotate differently) The more interesting question lies in Physics: DO gravity and electromagnetism share the same nature OR are they only isomorphic!? If two thing follow the same axioms they will follow the same theorems.
Nice way to solve. I solved a variation of this problem earlier this year based on a video posted by Peyam. There's an amazing detail contained in the problem. x² = 4² - [√(16-x²)]². x² = 3² - [√(9-x²)]². So 4² - [√(16-x²)]² =3² - [√(9-x²)]². Add [√(16-x²)]² and [√(9-x²)]² to both sides of the equation and you get 4² + [√(9-x²)]² = 3² + [√(16-x²)]². And so you obtain a variation of a pythagorean quadruple, a² + b² = c² + d². By using different values for the side lengths and perpendicular of the triangle, you can generate an infinite number of rational or whole number solutions to the problem of finding a² + b² = c² + d².
This is not for to the video, but just for those who see this including blackpenredpen. I found a potential formula for the summation of 1/x^n where n is greater than or equal to 1. d^n / d r^n { ln[ (x+r-1)! / (r-1)! ] }/n! with r being a constant equaling 1. ( Note I used d for the partial derivative of r).
You learn to love geometric representations in engineering - on the engineering HSC (Australia(NSW) Higher School Certificate) exam it's valid to solve force diagram problems either analytically, or by drawing the diagram to scale and measuring. If you're neat enough, it's a lot quicker, and about as accurate.
The Solution was pretty impressive . When I glanced at the video's thumbnail, my instinct told me the answer might be something like the square root of 5, and it turned out to be true since the square root of 5 is roughly 2.4. Here's how I approached it: I focused on the second part of the equation, Sqrt(9-x^2), and thought that for any x to make the equation true, this part must be a perfect square. Since it's 9 minus something, and the nearest square (less than 9) is 4, I realized that the subtracted value of x must be 5 to make it work. Voila! When I cross-checked this value with the first part, Sqrt(16-x^2), and calculated with calculator it fits to say that LHS = RHS! :)
for quadratic (max value of exponent is 2) equations as we can see here, theres a maximum of 2 solutions, including complex and real solutions here, theres 2 solutions an both are real, therefore there are no imaginary/complex solutions generally, a polynomial equation with a maximum exponent/power of "n", has upto "n" solutions, including both real and complex solutions.. so here theres no complex roots
@@TazwaarAhmed09 Just because it has x^2 max term doesn't mean an equation like that is quadratic. Equation in the video is not quadratic. Also not polynomial.
I thought about an interesting question and I couldn't solve it. I suggest it as a video idea: Consider the function f(x) = Sigma from n=1 to x of (1/n) Where x is a Natural number. Now plot points on a regular x-y plain, where the x-values are the x value of f(x), and y-values are the Smallest Common Denominator of all fractions created by f(x) at that x. In short, a coordinate is: (x, SCD). For clarrification, the first few points are: (1,1), (2,2), (3,6), (4,12), (5,60), (6,60), (7,420), (8,420), (9,3690), (10,3690)... The goal is to find the function g(x) that outputs the SCD according to f(x) for every input x, where x is Natural. Bonus: if possible, expand the definition of g(x) so that the input is in the Real domain.
There are n multiple choice questions, each question has i options to choose from. Step 1: Randomly choose the mth option (with m less than or equal to i and m greater than or equal to 1) in the first multiple choice question Step 2: Repeat the option in the 1st multiple-choice question in the next (k-1) multiple-choice questions. Step 3: To choose the option in the (k+1) multiple choice question, we will choose in the following way for each case: Case 1: If the option chosen in the kth multiple choice question is the mth option (with m smaller than i), then choose the (m+1)th option. Case 2: If the option chosen in the kth multiple choice question is the ith option, then choose the 1st option. Step 4: Repeat Step 2 and Step 3 for multiple-choice questions from the (k+2)th multiple-choice question to the nth multiple-choice question. Each multiple choice question has only 1 correct answer. Let t be the number of multiple-choice questions answered correctly in n multiple-choice questions, t follows the Bernouli distribution. Find k to t max.
Q: can you solve x^4+x^2+1=0 by using the quadratic formula? . . . Ans: yes (x^2)^2+x^2+1=0, and now we have a quadratic equation in terms of x^2 x^2=-1+-sqrt of 1^2-4(1)(1)/2(1) x^2=-1+-isqrt of 3/2, and now take the square root of both sides, we get: x=+-sqrt of( -1+-isqrt of 3/2), and now we can square both the numenator and the denominator, we get: x=+-i+-isqrt of 3/2 (x is equal to plus or minus i plus or minus the square root of 3 all over 2)
√(16-x²) + √(9-x²) = 5 √(4²-x²) + √(3²-x²) = 5 Notice that there's a Pythagorean Theorem thing going on Draw one right triangle with one side being x long, and the hypotenuse 4 long. The bottom side is unknown. Draw another right triangle, with the hypotenuse being 3 long this time. The bottom side is, again, unknown. Connect the triangles at the x long side. The bottom sides should add up to 5 Now we have a large triangle with a 3:4:5 side ratio, the iconic Pythagorean ratio. The 3 long and 4 long sides are perpendicular. Use the big triangle's area to calculate x. Notice that it is x tall. 3 • 4 / 2 = 5x / 2 x = 12 / 5 x = 2.4
Just noticed that even x = 0 as well as x= +- 12/5 Since, we are using the 3-4-5 triplet to find the results The one of the trivial solutions to the problem!! Awesome explanation!!!! Looking forward to more such videos
1: If the resulting triangle isn't a right triangle, we can use Heron's formula. 2: Haha, ChatGPT did it wrong again! ChatGPT is not good at math, but quite good at programming!
On a recent StarTalk podcast episode, Prof. Charles Liu said on his (presumably at-home) exams, he asks his students to use chaptgpt to solve a problem, and then to explain why it's wrong 😂
Even without the perfect numbers leading to the second right triangle, Heron's formula can be used to get the second relation for the area: let semiperimeter s = (a+b+c)/2 then area A = sqrt[ s(s-a)(s-b)(s-c) ]. In this problem s is 6 and A is also 6.
Yeah, square, cancel terms, square, and you are done. If the numbers were different and stuff didn't cancel nicely we'd have had to solve a quadratic equation along the way (scary!), but we'd still get a solution. I can't help but feel like there is a very wrong lesson in this, you can't count on always getting nice equations in the wild, looking for an elegant way, rather than just a way, to solve something is most often a waste of time.
I used rationalisation and it also works here. On rationalising (16-x²)½+(9-x²)½=5, We get (16-x²)½-(9-x²)½=7/5. Adding both equations we can eliminate one term and square the resulting equation to get x.
Hi Blackpenredpen, I have two things to tell you! First of all, could you make the 100 integrals file (the first video) freely available? Secondly, I offer you a huge integral! The integral of 1/(1+x^5). Otherwise good luck! Bye !
Hey there, Steve. Since I’m in Info Tech, I also had to learn CS (computer science). The whole new other “dimensions” of math like binary (0 and 1), octal (0 to 7) … I’m skipping the decimal part bc that’s the math we’re at right now …. Hexadecimal (base-16 that’s 0 to 9 and A to F representing 10 to 15. Like 1F for example …. That’s 31 in decimal and so forth. You get the gist), base-32, base-64 and finally, base-85. I learned there’s a multiplication table of hexadecimals. Google it up all you want. Now! I’m curious. Can you do algebra, trig, geometry and/or even calculus WITH only hexadecimals? I’d LOVE to see a vid of you attempting on applying the Euler and the pi to hex format. Hehe. Pretty please?! Can you do that?! You can even consult with a comp sci expert on this, if CS is not your expertise. Wink.
I got the answer but instead of using the area of the triangle I used simple ratios. Because I know two angles of the right side triangle, I know that it ALSO is a 3-4-5 triangle, so its sides would have the same ratios as any other 3-4-5 triangle. That gave 3/x=4/5. Very very easy to calculate.
Very nice solution! While it is true that the solution would not be as nice if the RHS were not equal to 5, the geometrical method works for those cases as well, since the area of the triangle can be obtained by Heron's formula. For RHS = 6, the answer is +/-sqr(455)/12.
Can you make a video about this interesting problem--> If z = x + iy is a complex number where x and y are integers. Then, the area of the rectangle whose vertices are the roots of the equation----> w z^3 + z w^3 = 350, where w=conjugate of z
Wouldnt the negative still be extraneous since it's asking for a measurement? Yes negative measurements can exist but it doesn't really make sense in this case
Well the original question doesn't ask for a measurement it just asks for the value of x, he just used geometrical terms to solve this equation in another manner, obviously if you use geometry you will only deal with positive values, but since the original question is algebraic we can also have negative values for x
For this to work the larger triangle has to be a right triangle. No, I was wrong here. Consider a 13-14-15 triangle. Used Heron’s formula to get area=84 and height above 15 side is 11.2. Sqrt(169-x^2)+sqrt(196-x^2)=15, x=+/-11.2. This is great.
Respected sir, Find the value of the following questions geometrically in the principle value range: (a) arcsin(sin 23) (b) arcsin(sin24) Where both 23 and 24 are in radian.
I just multiplied both sides to a conjugate sqrt(16-x^2)-sqrt(9-x^2) that resulted in sqrt(16-x^2)-sqrt(9-x^2)=1.4 then I simply used the Gauss method and summarized this equation with the source one so 2sqrt(16-x^2)=6.4 => sqrt(16-x^2)=3.2 => 16-x^2=10.24 => x^2=5.76 => x=+-2.4
There's also a nice algebraic solution: Multiply by the conjugate nd we get 7/5=√(16-x²) -√(9-x²) Then we add the first eq nd divide by 2 we get √(16-x²)=16/5 nd then we end up gettin the same result 👩🏻🎤
Hi BlackPenRedPen! Could you please make a video about solving: i ^ x = x ? I have tried it, but I keep arriving at the wrong solution, and I cannot find explanations online. I tried to get the maximum amount of complex solutions by using all branches of Lambert W Function and using all possible theta values to represent i as e^(i*theta). But it is wrong on WolframAlpha. Can you please help?
I got this: i^x=x, x=e^i(pi/2+2piN)x, (where N is an integer) x=cos((pi/2+2piN)x)+isin((pi/2+2piN)x). if x=a+bi: a+bi = cos((pi/2+2piN)a+(pi/2+2piN)bi)+isin((pi/2+2piN)a+(pi/2+2piN)bi) I'll say M=pi/2+2piM to make things shorter, so a+bi=cos(Ma+Mbi)+isin(Ma+Mbi) a+bi=cos(Ma)cos(Mbi)-sin(Ma)sin(Mbi)+isin(Ma)cos(Mbi)+icos(Ma)sin(Mbi) a+bi=cos(Ma)cosh(Mb)-isin(Ma)sinh(Mb)+isin(Ma)cosh(Mb)-cos(Ma)sinh(Mb) now you can make a system with the real and imaginary parts: a=cos(Ma)cosh(Mb)-cos(Ma)sinh(Mb) b=sin(Ma)cosh(Mb)-sin(Ma)sinh(Mb) a=cos(Ma)(cosh(Mb)-sinh(Mb)) b=sin(Ma)(cosh(Mb)-sinh(Mb)) a=cos(Ma)*e^-(Mb) b=sin(Ma)*e^-(Mb) a*e^(Mb)=cos(Ma) b*e^(Mb)=sin(Ma) from the first equation, b=ln(cos(Ma)/a)/M dividing the second equation by the first, b=a*tan(Ma) putting them together, Ma*tan(Ma)=ln(cos(Ma)/a) and once you have a solution for a, to find b you just use b=a*tan(Ma) plotting with desmos and looking for approximations gave me approximate answers that work but I don't know how to get exact answers. Wolframalpha's answer seems a bit suspicious to me since it fails to give even an approximate answer, and is unable to calculate W(-i*pi/2), which it says is part of the answer, but I don't know enough to know whether what wolfram is saying is right.
before watching the video: this can be rewritten as sqrt(16-x^2)=5-sqrt(9-x^2) These are the equations of semicircles, so basically the intersections between a circle of radius 4 with center in (0,0) and a circle of radius 3 with center in (5,0). These form 3-4-5 right triangles, so the distance of the intersection from the hypotenuse is 3*4/5=12/5. The solutions are x=12/5 and x=-12/5
You can use this method: Suppose we have a right triangle where AB^2 + AC^2 = BC^2 and BC = BD + DC So the height AD which is x in this video's case will be AD ^ 2 = BD * DC And then you can solve the problem. Right, isn't it?
what was up with holding that red white and black stuffed looking thing in the past? I was curious when i saw your past vids. Interesting content btw! Very creative, different, and analytical.
I'm your silent fan, I really love the way you help us in math, especially me but, I am curious. How did you become master at math? how much time do you solve math before? can you make a video for it?
Get started with a 30-day free trial on Brilliant: 👉brilliant.org/blackpenredpen/ ( 20% off with this link!)
I’M JOSEPH STALIN’S MIGHTY FROG
Your "phd friend" video was hilarious, I was laughing about that for hours. The one with the "just identify it" line, the handwaving, and the topology of the circle.
:D
Hey bprp! Can you do this math equation? It was in my test.
sin(cos(tan(x)))=1
But base on the video, you could use the pythagorams inverse formula, also you have on you channel!!!
It would be more intresting!!
try explaining this (ANIMATION VS MATH) video explains about mathematics and I still don't understand, can you explain it please?
I think this is a spot for sybermath's substitution trick.
Let a = sqrt(16 - x^2). Let b = sqrt(9 - x^2). Then, we are given that a + b = 5. Also, from the definitions of a and b, we have a^2 = 16 - x^2, and b^2 = 9 - x^2. Subtracting those, we get a^2 - b^2 = 7. So, (a + b)(a - b) = 7. But, since a + b = 5, we get 5(a - b) = 7, or a - b = 1.4. Add that to a + b = 5, and we get 2a = 6.4, or a = 3.2. Plugging that back in to a = sqrt(16 - x^2). 3.2 = sqrt(16 - x^2), so 10.24 = 16 - x^2. x^2 = 16 - 10.24 = 5.76 = 2.4^2. So, x = +/- 2.4.
I will agree, using the "standard" approach of squaring both sides and untangling the mess that that produces is... a giant pain. But, sybermath's substitution trick avoids that nicely.
Nice,i like sybermath trick too,but no need for any tricks here.just square sqrt(16-x^2)-5 and -sqrt(9-x^2) and you get after simple algebra x^2=5.76 so x=+-2.4
I squared both sides (25=16+9-2x^2+2√((9-x^2)(16-x^2))) after that u get x^2=√((16-x^2)(9-x^2))
Then squaring both sides u will end up 25x^2=16×9
Solving this simple equation u get postive or negative 12/5
@@abdelrahmanayman8415 If you are going to square both sides... Subtract one of them first.
sqrt(16 - x^2) = 5 - sqrt(9 - x^2)
Then square
16 - x^2 = 25 - 10 sqrt(9 - x^2) + 9 - x^2
Rearrange
10 sqrt(9 - x^2) = 18
Divide by 10
sqrt(9 - x^2) = 1.8
Square
9 - x^2 = 3.24
Rearrange
x^2 = 5.76
Then solve for x
x = 2.4 or -2.4
Notice that you don't have to deal with as big an equation as you did.
sybermath's trick? What makes this trick to be "sybermath's trick"?
That is so fucking cool, thank you so much for sharing this!?!??
"Trust in Geometry" people when meet "i"
Probably a lot of important questions (like up to 3rd deg polynomials) are geometry related.
This is a very interesting question 🙋
Euclidean geometry is the future 😆
You can always construct a geometric figure that represents an algebraic relationship. That is what makes analysis so beautiful.
@@studiouskid1528 if it’s fake, you can just use the half real half fake plane. The noobs call it complex but I’ve always thought it was simple.
@@studiouskid1528 2 real axes and one imaginary axis. x->z = a+bi.
Now I will use geometry whenever I can to solve funky algebraic equations.
That ChatGPT attempt 💀💀
“Ai will take over the world!”
Meanwhile, AIs: “tHrEe pLuS fOuR eQuAlS fIvE”
It was ChatGPT3.5, not ChatGPT4, so of course it's not going to be good at math.
@@Macieks300 you say chat GPT 4 is good at math then
@@Macieks300 gpt 4 is equally garbage at math lmao
@@saj_f0 Yeah, it's way better at math.
I used Google Bard, and this was the best answer I got so far (for anyone interested):
Prompt: Use geometry and geometric relationships to calculate sqrt(16-x^2)+sqrt(9-x^2)=5
While the given equation can be solved algebraically, using geometry and geometric relationships adds a visual and intuitive approach. Here's how we can solve it:
1. Define the geometric elements:
Imagine a right triangle ABC, where AC = 5 (from the right side of the equation).
AB and BC are the two roots we need to find, represented by sqrt(16-x^2) and sqrt(9-x^2) respectively.
Point D lies on AC such that AD = sqrt(16-x^2) and DC = sqrt(9-x^2).
2. Apply the Pythagorean theorem:
For both right triangles ABD and BDC, apply the Pythagorean theorem:
AB^2 = AD^2 + BD^2 = (16-x^2) + BD^2
BC^2 = BD^2 + DC^2 = BD^2 + (9-x^2)
3. Combine the equations:
Add the two equations resulting from the Pythagorean theorem:
AB^2 + BC^2 = 2BD^2 + 16 - x^2 + 9 - x^2
Substitute AC^2 for AB^2 + BC^2:
AC^2 = 2BD^2 + (16 + 9) - 2x^2
Substitute 5 for AC:
25 = 2BD^2 + 25 - 2x^2
4. Solve for BD^2:
Simplify the equation:
2BD^2 = 2x^2
BD^2 = x^2
5. Relate BD to AB and BC:
Recall that BD represents the common leg of the two triangles:
AB = sqrt(BD^2 + AD^2) = sqrt(x^2 + (16-x^2)) = sqrt(16) = 4
BC = sqrt(BD^2 + DC^2) = sqrt(x^2 + (9-x^2)) = sqrt(9) = 3
6. Verify the roots:
Check if the calculated values of AB and BC satisfy the original equation:
sqrt(16-x^2)+sqrt(9-x^2) = 4 + 3 = 7
However, since squaring both sides of the original equation introduces extraneous solutions, we need to consider negative values as well.
7. Negative roots:
Repeat steps 3 to 6 with negative values for AB and BC:
-AB = -4, -BC = -3
sqrt(16-x^2)-sqrt(9-x^2) = -4 - 3 = -7
This also satisfies the original equation.
Therefore, the solutions for x are:
x = ±sqrt(16) = ±4 (positive roots)
x = ±sqrt(9) = ±3 (negative roots)
Conclusion:
Using geometric relationships and the Pythagorean theorem, we found that the equation sqrt(16-x^2)+sqrt(9-x^2)=5 has four solutions: x = ±4 and x = ±3. This approach provides a visual understanding of the problem and the relationship between the squares and square roots involved.
Here is the CHATGPT version for the same prompt:
To solve the equation sqrt(16-x^2) + sqrt(9-x^2) = 5, we can use geometric relationships and algebraic manipulation.
Let a = sqrt(16 - x^2) and b = sqrt(9 - x^2). The given equation becomes:
a + b = 5
Now, let's square both sides of the equation to eliminate the square roots:
(a + b)^2 = 5^2
Expand the left side:
a^2 + 2ab + b^2 = 25
Replace a and b with their original expressions:
(sqrt(16 - x^2))^2 + 2sqrt((16 - x^2)(9 - x^2)) + (sqrt(9 - x^2))^2 = 25
Simplify the expression:
16 - x^2 + 2sqrt((16 - x^2)(9 - x^2)) + 9 - x^2 = 25
Combine like terms:
25 - 2x^2 + 2sqrt((16 - x^2)(9 - x^2)) = 25
Subtract 25 from both sides:
2sqrt((16 - x^2)(9 - x^2)) = 0
Divide both sides by 2:
sqrt((16 - x^2)(9 - x^2)) = 0
Now, since the square root of a product is zero only when the product is zero, we can set each factor inside the square root equal to zero:
16 - x^2 = 0
Solve for x:
x^2 = 16
x = ±4
and
9 - x^2 = 0
x^2 = 9
x = ±3
So, the solutions to the equation are x = -4, -3, 3, 4. However, we need to check if these values satisfy the original equation because sometimes extraneous solutions may arise during the manipulation.
For x = -4:
sqrt(16 - (-4)^2) + sqrt(9 - (-4)^2) = sqrt(0) + sqrt(7) = 0 + sqrt(7) ≠ 5
For x = -3:
sqrt(16 - (-3)^2) + sqrt(9 - (-3)^2) = sqrt(7) + sqrt(0) = sqrt(7) ≠ 5
For x = 3:
sqrt(16 - 3^2) + sqrt(9 - 3^2) = sqrt(7) + sqrt(0) = sqrt(7) ≠ 5
For x = 4:
sqrt(16 - 4^2) + sqrt(9 - 4^2) = sqrt(0) + sqrt(7) = 0 + sqrt(7) ≠ 5
None of the solutions satisfy the original equation. Therefore, there are no real solutions to the given equation.
@@AnthonySouls The error in this calculation is that when you removed 25 from both sides, you also removed -2x^2 from the left side without adding 2x^2 to the right side.
Great simple geometric solution 😊
At 3:30, getting x is easier by similar triangles. Each small triangle is similar to the large one by AA theorem, as they share an angle and also have a right angle. Using the left triangle, you get x/4=3/5, and on the right you get x/3=4/5. In either case, multiply by the denominator on the left to get x=12/5.
I'm loving this series of yours, the "Believe in __ not ____". Could we get a trigonometry edition?
Great idea! I will have to think about it.
Also there exist inverse pythagoran where 1/a^2 + 1/b^2 = 1/d^2 so in our case it would be : 1/4^2 + 1/3^2 = 1/x^2 and it would be equal to this anwswer
1. Firstly, we should determinate a solution space, if we working with real numbers:
(16 - x² ≥ 0) & (9 - x² ≥ 0);
((4 - x) × (4 + x) ≥ 0) & ((3 - x) × (3 + x) ≥ 0);
(- 4 ≤ x ≤ 4) & (- 3 ≤ x ≤ 3);
- 3 ≤ x ≤ 3.
Last string is the solution space of our equation in real numbers.
2. Also we can use geometric mean formula, because we have an altitude from right angle in rectangular triangle ("x" is the length of it):
x = sqrt (sqrt (16 - x²) × sqrt (9 - x²));
x = sqrt (sqrt ((16 - x²) × (9 - x²));
x⁴ = (16 - x²) × (9 - x²);
x⁴ = 16 × 9 - 16 x² - 9 x² + x⁴;
x⁴ = 16 × 9 - 25 x² + x⁴;
0 = 16 × 9 - 25 x²;
25 x² = 16 × 9;
25 x² - 16 × 9 = 0;
(5 x)² - (4 × 3)² = 0;
(5 x - 4 × 3) • (5 x + 4 × 3) = 0;
1) 5 x - 4 × 3 = 0;
5 x = 4 × 3;
x = 4 × 3 / 5;
x = 12 / 5;
x = 2,4.
2) 5 x + 4 × 3 = 0;
5 x = (- 4) × 3;
x = (- 4) × 3 / 5;
x = - 12 / 5;
x = - 2,4.
3. Both of values of "x" entered in our solution space, because - 3 ≤ - 2,4 ≤ 2,4 ≤ 3 - is right inequality, that is why answer is: x = ± 2,4 (in real numbers).
It is sort of isomorphic transform I think. Like Laplace or Fourier transfom. Many things are isomorphic even when they are very different.
Isomorphisms!
don't think that it is an isomorphism at least not in the usual sense.
@@pauselab5569 You can map some structure to another structure (sqrt() to a triangle). Map some operation to another operation (addition to position in space). Then you should unmap the result.
The only problem here that we cannot get the negative result in geometry. We cannot be sure there no other colutions using this mapping. That is true.
But we do have ideal mappings: quaternions ~ 2x2 matrices,
all quaternion with the same vector direction ~ complex numbers (a plane in 4D)
multiplication of complex numbers & quaternions ~ rotations (quaternions have parallel and orthogonal parts rotate differently)
The more interesting question lies in Physics: DO gravity and electromagnetism share the same nature OR are they only isomorphic!? If two thing follow the same axioms they will follow the same theorems.
What ... are you talking about?
@@samueldeandrade8535isomorphisms. Get with the program
Nice way to solve. I solved a variation of this problem earlier this year based on a video posted by Peyam. There's an amazing detail contained in the problem. x² = 4² - [√(16-x²)]². x² = 3² - [√(9-x²)]². So 4² - [√(16-x²)]² =3² - [√(9-x²)]². Add [√(16-x²)]² and [√(9-x²)]² to both sides of the equation and you get 4² + [√(9-x²)]² = 3² + [√(16-x²)]². And so you obtain a variation of a pythagorean quadruple, a² + b² = c² + d². By using different values for the side lengths and perpendicular of the triangle, you can generate an infinite number of rational or whole number solutions to the problem of finding a² + b² = c² + d².
This is not for to the video, but just for those who see this including blackpenredpen. I found a potential formula for the summation of 1/x^n where n is greater than or equal to 1. d^n / d r^n { ln[ (x+r-1)! / (r-1)! ] }/n! with r being a constant equaling 1. ( Note I used d for the partial derivative of r).
It is actually d/dr {ln[ d^(n-1) / dr^(n-1) (x+r-1)!/(r-1)!]} and the bounds of summation are 1 and infinity.
Really cool problem! Thank you as always
i used angles to find the value of x in the end , assume one angle tetha rest are either 90-tetha or tetha which makes it much easier
This is gold! Thank you for such a useful video!
I have always loved the geometrical representation of equations (mostly from 3blue 1brown) but this method is just ♥
You learn to love geometric representations in engineering - on the engineering HSC (Australia(NSW) Higher School Certificate) exam it's valid to solve force diagram problems either analytically, or by drawing the diagram to scale and measuring. If you're neat enough, it's a lot quicker, and about as accurate.
*Standing Ovation!*
Always lovely to see elegance in math problems like this. :)
This is great. The numbers work very well. Thank you for that.
Glad to see you again!
Nice solution!
ти ща де
The Solution was pretty impressive . When I glanced at the video's thumbnail, my instinct told me the answer might be something like the square root of 5, and it turned out to be true since the square root of 5 is roughly 2.4.
Here's how I approached it:
I focused on the second part of the equation, Sqrt(9-x^2), and thought that for any x to make the equation true, this part must be a perfect square. Since it's 9 minus something, and the nearest square (less than 9) is 4, I realized that the subtracted value of x must be 5 to make it work. Voila! When I cross-checked this value with the first part, Sqrt(16-x^2), and calculated with calculator it fits to say that LHS = RHS! :)
What about them complex roots
for quadratic (max value of exponent is 2) equations as we can see here, theres a maximum of 2 solutions, including complex and real solutions
here, theres 2 solutions an both are real, therefore there are no imaginary/complex solutions
generally, a polynomial equation with a maximum exponent/power of "n", has upto "n" solutions, including both real and complex solutions.. so here theres no complex roots
@@TazwaarAhmed09 Just because it has x^2 max term doesn't mean an equation like that is quadratic. Equation in the video is not quadratic. Also not polynomial.
He used a blue pen
wow
He escaped the "MATRIX".
Question) If Y={x such that x is a positive factor of the number [(2^p-1)(2^p - 1)] , where (2^p - 1) is a prime number}. Write Y in the rooster form
I thought about an interesting question and I couldn't solve it. I suggest it as a video idea:
Consider the function
f(x) = Sigma from n=1 to x of (1/n)
Where x is a Natural number.
Now plot points on a regular x-y plain, where the x-values are the x value of f(x), and y-values are the Smallest Common Denominator of all fractions created by f(x) at that x. In short, a coordinate is: (x, SCD).
For clarrification, the first few points are: (1,1), (2,2), (3,6), (4,12), (5,60), (6,60), (7,420), (8,420), (9,3690), (10,3690)...
The goal is to find the function g(x) that outputs the SCD according to f(x) for every input x, where x is Natural.
Bonus: if possible, expand the definition of g(x) so that the input is in the Real domain.
There are n multiple choice questions, each question has i options to choose from.
Step 1: Randomly choose the mth option (with m less than or equal to i and m greater than or equal to 1) in the first multiple choice question
Step 2: Repeat the option in the 1st multiple-choice question in the next (k-1) multiple-choice questions.
Step 3: To choose the option in the (k+1) multiple choice question, we will choose in the following way for each case:
Case 1: If the option chosen in the kth multiple choice question is the mth option (with m smaller than i), then choose the (m+1)th option.
Case 2: If the option chosen in the kth multiple choice question is the ith option, then choose the 1st option.
Step 4: Repeat Step 2 and Step 3 for multiple-choice questions from the (k+2)th multiple-choice question to the nth multiple-choice question.
Each multiple choice question has only 1 correct answer. Let t be the number of multiple-choice questions answered correctly in n multiple-choice questions, t follows the Bernouli distribution. Find k to t max.
That was a neat approach to this problem! When I read the title, I thought more about circles than right triangles.
I would never have thought of approaching the problem that way. Very interesting!
Thank you, so much 😊
Q: can you solve x^4+x^2+1=0 by using the quadratic formula?
.
.
.
Ans: yes
(x^2)^2+x^2+1=0, and now we have a quadratic equation in terms of x^2
x^2=-1+-sqrt of 1^2-4(1)(1)/2(1)
x^2=-1+-isqrt of 3/2, and now take the square root of both sides, we get:
x=+-sqrt of( -1+-isqrt of 3/2), and now we can square both the numenator and the denominator, we get:
x=+-i+-isqrt of 3/2
(x is equal to plus or minus i plus or minus the square root of 3 all over 2)
This was fun! thank you for your videos
√(16-x²) + √(9-x²) = 5
√(4²-x²) + √(3²-x²) = 5
Notice that there's a Pythagorean Theorem thing going on
Draw one right triangle with one side being x long, and the hypotenuse 4 long. The bottom side is unknown.
Draw another right triangle, with the hypotenuse being 3 long this time. The bottom side is, again, unknown.
Connect the triangles at the x long side. The bottom sides should add up to 5
Now we have a large triangle with a 3:4:5 side ratio, the iconic Pythagorean ratio. The 3 long and 4 long sides are perpendicular.
Use the big triangle's area to calculate x. Notice that it is x tall.
3 • 4 / 2 = 5x / 2
x = 12 / 5
x = 2.4
Im gonna use this question to show my friends that geometry is beautiful and not just shapes
Super interesting video!
Wow, Thank you, so amazing, I love this technique.
Just noticed that even x = 0 as well as x= +- 12/5
Since, we are using the 3-4-5 triplet to find the results
The one of the trivial solutions to the problem!!
Awesome explanation!!!!
Looking forward to more such videos
Except x=0 doesn't work. When x=0, sqrt(16-x^2)=4 and sqrt(9-x^2)=3. But 4+3=7, and 7≠5.
we can also able to do this in another way:
✓16-x²=(5-✓9-x²)
then we can square both sides
it will be very easy
This will be a very good watch!
1: If the resulting triangle isn't a right triangle, we can use Heron's formula.
2: Haha, ChatGPT did it wrong again! ChatGPT is not good at math, but quite good at programming!
On a recent StarTalk podcast episode, Prof. Charles Liu said on his (presumably at-home) exams, he asks his students to use chaptgpt to solve a problem, and then to explain why it's wrong 😂
I was hoping you’d use the inverse pythagorean theorem 1/a^2 + 1/b^2 = 1/h^2 then you get 1/x^2 = 25/144 so x = +/- 12/5
Even without the perfect numbers leading to the second right triangle, Heron's formula can be used to get the second relation for the area: let semiperimeter s = (a+b+c)/2 then area A = sqrt[ s(s-a)(s-b)(s-c) ]. In this problem s is 6 and A is also 6.
After making the triangle, instead of equations areas you can use the "inverse" Pythagoras theorem ie 1/x^2 = 1/3^2+1/4^2
Try this:
1+power(2+power(3+…+power(n+…,1/n),1/3),1/2)=?where n is natural and goes to infinity.
I did this one the brute force way, and it was surprisingly not too bad. It was a miracle of canceling terms so that it solved pretty neatly.
Yeah, square, cancel terms, square, and you are done. If the numbers were different and stuff didn't cancel nicely we'd have had to solve a quadratic equation along the way (scary!), but we'd still get a solution. I can't help but feel like there is a very wrong lesson in this, you can't count on always getting nice equations in the wild, looking for an elegant way, rather than just a way, to solve something is most often a waste of time.
@@eBusinessjma Sure; by all means take a minute to look for an inspired way to approach a problem, but don't expect there will be one.
Thank you, so much
As soon as I saw the problem ,I knew what you were going to do
I used rationalisation and it also works here. On rationalising (16-x²)½+(9-x²)½=5,
We get (16-x²)½-(9-x²)½=7/5.
Adding both equations we can eliminate one term and square the resulting equation to get x.
Imperial!
As an math aspirant i will use it
May i ask since n[sin(180degree/n)] helps u calculate pi. If lim n--> infinity, is n(sin(180degree/n)) = pi?
anyone answer me pls ty
Great solution!
If it is not right angled triangle we can use 1/2 ab sin(x) to find the area
Wonderful solution!
Trust in algebra too
Let 9-x² = y²
=> x² = 9-y²
Plugging it in our equation, we get
√(7+y²) + y = 5
7 + y² = (5 - y)²
7 + y² = y² - 10y + 25
10y = 18
y = 9/5
Now,
x²=9-y²
=9-(9/5)²
=144/25
x=±12/5
Brilliant, I did not think to use build the second triangle in that way. I made a mental calculation and found sqr 7 is a good solution😊.
I always love having a look inside!
La longitud de la altura referida a la hipotenusa en un triángulo rectángulo es igual al producto de los catetos dividido por la hipotenusa, x=(3*4)/5
4:24 I would have used Heron's formula for the area of the triangle here and equate it with (5×)/2 😂😂
Hi Blackpenredpen, I have two things to tell you! First of all, could you make the 100 integrals file (the first video) freely available? Secondly, I offer you a huge integral! The integral of 1/(1+x^5). Otherwise good luck! Bye !
Very good teaching. Thank you !
I never even thought about it before, yet it just seem intuitive enough to work soo well. Thank you for the educational video.
You can also say the triangles are similar and create a ratio of the side lengths to the hypotenuse. 4/5 = x/3
Can you try the integral of x/sinh(x) from -∞ to ∞? It comes out to a really nice answer, but I have no idea how you’d actually solve it…
老师可以讲一下怎么积分Tan(cos(x)) 从0 到2pie吗
Hey there, Steve. Since I’m in Info Tech, I also had to learn CS (computer science). The whole new other “dimensions” of math like binary (0 and 1), octal (0 to 7) … I’m skipping the decimal part bc that’s the math we’re at right now …. Hexadecimal (base-16 that’s 0 to 9 and A to F representing 10 to 15. Like 1F for example …. That’s 31 in decimal and so forth. You get the gist), base-32, base-64 and finally, base-85.
I learned there’s a multiplication table of hexadecimals. Google it up all you want.
Now! I’m curious. Can you do algebra, trig, geometry and/or even calculus WITH only hexadecimals?
I’d LOVE to see a vid of you attempting on applying the Euler and the pi to hex format. Hehe. Pretty please?! Can you do that?! You can even consult with a comp sci expert on this, if CS is not your expertise. Wink.
Damn...I was always an algebra guy but now I think geometry also kicks. Thanks sir.
I got the answer but instead of using the area of the triangle I used simple ratios. Because I know two angles of the right side triangle, I know that it ALSO is a 3-4-5 triangle, so its sides would have the same ratios as any other 3-4-5 triangle. That gave 3/x=4/5. Very very easy to calculate.
Very nice solution! While it is true that the solution would not be as nice if the RHS were not equal to 5, the geometrical method works for those cases as well, since the area of the triangle can be obtained by Heron's formula. For RHS = 6, the answer is +/-sqr(455)/12.
very interesting approach!
Can you make a video about this interesting problem--> If z = x + iy is a complex number where x and y are integers. Then, the area of the rectangle whose vertices are the roots of the equation----> w z^3 + z w^3 = 350, where w=conjugate of z
0:53.............................ya don't believe in squaring both the sides but believe in taking square root on both sides.... 😌
Euclid's Height Theorem: h² = pq
Euclid's Leg Theorem: a² = cp, b² = cq
Wouldnt the negative still be extraneous since it's asking for a measurement? Yes negative measurements can exist but it doesn't really make sense in this case
Well the original question doesn't ask for a measurement it just asks for the value of x, he just used geometrical terms to solve this equation in another manner, obviously if you use geometry you will only deal with positive values, but since the original question is algebraic we can also have negative values for x
Congratulation! It is fantastic!!!
For this to work the larger triangle has to be a right triangle.
No, I was wrong here. Consider a 13-14-15 triangle. Used Heron’s formula to get area=84 and height above 15 side is 11.2.
Sqrt(169-x^2)+sqrt(196-x^2)=15, x=+/-11.2. This is great.
Respected sir,
Find the value of the following questions geometrically in the principle value range:
(a) arcsin(sin 23)
(b) arcsin(sin24)
Where both 23 and 24 are in radian.
Even though substituting 9 - x^2 as M might be a better method, this is a very interesting!!
I just multiplied both sides to a conjugate sqrt(16-x^2)-sqrt(9-x^2) that resulted in sqrt(16-x^2)-sqrt(9-x^2)=1.4 then I simply used the Gauss method and summarized this equation with the source one so 2sqrt(16-x^2)=6.4 => sqrt(16-x^2)=3.2 => 16-x^2=10.24 => x^2=5.76 => x=+-2.4
with similar triangles you can see the answers with ratios (either sub triangle works), 5 : 3 as 4 : x implies x=12/5
Lol.... ChatGPT really bombed that question.
I know it has to do with theorem of Pythagoras. But I am still thinking. Excellent question.
There's also a nice algebraic solution:
Multiply by the conjugate nd we get 7/5=√(16-x²) -√(9-x²)
Then we add the first eq nd divide by 2 we get √(16-x²)=16/5 nd then we end up gettin the same result 👩🏻🎤
Can we integrate cosx*e^(-cosx)
Pls reply
Hi BlackPenRedPen! Could you please make a video about solving: i ^ x = x ? I have tried it, but I keep arriving at the wrong solution, and I cannot find explanations online. I tried to get the maximum amount of complex solutions by using all branches of Lambert W Function and using all possible theta values to represent i as e^(i*theta). But it is wrong on WolframAlpha. Can you please help?
I got this:
i^x=x,
x=e^i(pi/2+2piN)x, (where N is an integer)
x=cos((pi/2+2piN)x)+isin((pi/2+2piN)x).
if x=a+bi:
a+bi = cos((pi/2+2piN)a+(pi/2+2piN)bi)+isin((pi/2+2piN)a+(pi/2+2piN)bi)
I'll say M=pi/2+2piM to make things shorter, so
a+bi=cos(Ma+Mbi)+isin(Ma+Mbi)
a+bi=cos(Ma)cos(Mbi)-sin(Ma)sin(Mbi)+isin(Ma)cos(Mbi)+icos(Ma)sin(Mbi)
a+bi=cos(Ma)cosh(Mb)-isin(Ma)sinh(Mb)+isin(Ma)cosh(Mb)-cos(Ma)sinh(Mb)
now you can make a system with the real and imaginary parts:
a=cos(Ma)cosh(Mb)-cos(Ma)sinh(Mb)
b=sin(Ma)cosh(Mb)-sin(Ma)sinh(Mb)
a=cos(Ma)(cosh(Mb)-sinh(Mb))
b=sin(Ma)(cosh(Mb)-sinh(Mb))
a=cos(Ma)*e^-(Mb)
b=sin(Ma)*e^-(Mb)
a*e^(Mb)=cos(Ma)
b*e^(Mb)=sin(Ma)
from the first equation, b=ln(cos(Ma)/a)/M
dividing the second equation by the first, b=a*tan(Ma)
putting them together, Ma*tan(Ma)=ln(cos(Ma)/a) and once you have a solution for a, to find b you just use b=a*tan(Ma)
plotting with desmos and looking for approximations gave me approximate answers that work but I don't know how to get exact answers.
Wolframalpha's answer seems a bit suspicious to me since it fails to give even an approximate answer, and is unable to calculate W(-i*pi/2), which it says is part of the answer, but I don't know enough to know whether what wolfram is saying is right.
before watching the video: this can be rewritten as sqrt(16-x^2)=5-sqrt(9-x^2) These are the equations of semicircles, so basically the intersections between a circle of radius 4 with center in (0,0) and a circle of radius 3 with center in (5,0). These form 3-4-5 right triangles, so the distance of the intersection from the hypotenuse is 3*4/5=12/5. The solutions are x=12/5 and x=-12/5
You can use this method:
Suppose we have a right triangle where AB^2 + AC^2 = BC^2 and BC = BD + DC
So the height AD which is x in this video's case will be AD ^ 2 = BD * DC
And then you can solve the problem.
Right, isn't it?
Please, I want you to create a formula to solve the equation
ax^3+bx^2+cx+b
And this one too
ax^3+bx=n
Look up cardano formula
Nice approach.
can this be generalized for solutions of the form: sqrt(a^2 - x^2) + sqrt(b^2 - x^2) = c where a,b,c are a recognized pyth. triple?
I don't know why this is made so complicated
It is clearly x=0
How do we do a multivariable limit using epsilon-delta method?
would you be able to do a video explaining how to solve equations such as 2^x + 3^x = 2 or 2^x + 5^x = 1 and other similar cases?
Любопытно! Хорошо бы ещё провести проверку 😘
what was up with holding that red white and black stuffed looking thing in the past? I was curious when i saw your past vids. Interesting content btw! Very creative, different, and analytical.
That was a cool problem Thank you 😊
I'm your silent fan, I really love the way you help us in math, especially me but, I am curious. How did you become master at math? how much time do you solve math before? can you make a video for it?
Well done, Alan!
@blackpenredpen Can you please do a break down of the integral arctan(x)/x^3 dx? I’m fascinated by it.
I f**ing love this guy!