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A Radically Radical Equation | See the Trick?
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- Опубліковано 20 жов 2023
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When solving the equation, we did not refer anywhere to the fact that x is real. It could also be complex, and we also get the only possibility that z = -4, which is not a solution. Namely, -5 is not equal to 5, but (-5)² is equal to 5².
3:30 This term can be estimated alot sharper:
For all real values of x, we have
x^2 >= 0
x^2 + 9 >= 9
sqrt(x^2 + 9) >= 3
1 + sqrt(x^2 + 9) >= 4.
But for x >= 4, we get
x^2 >= 16
x^2 + 9 >= 25
sqrt(x^2 + 9) >= 5
1 + sqrt(x^2 + 9) >= 6.
And so forth. So the sequence of x values is increasing and does not converge to a "fixpoint solution".
is not enough proof.
Nice, ple do trigonometry problems next
1 + sqrt(x^2 + 9) > sqrt(x^2 + 9) > sqrt(x^2) = |x| >= x, or, to collapse the inequalities, 1 + sqrt(x^2 + 9) > x for all values of x.
There is no solution.
As for people letting x range outside the real numbers, the problem is that the sqrt radical function is only defined on non-negative real numbers. sqrt(x) doesn't mean "one of the two numbers that square to equal x". It's a function and thus must be single-valued.
👍
x = 1 + √(x^2+9)
solutions we get is -4
putting x =(-4) LHS
x = -4
putting x=(-4) RHS
1+√((-4)^2+9)=
1+√(16+9)=
1+√(25)=
1-5(√a give +ve and -ve both value)
=-4
why did you not do this ?
Last time I checked, x² was always less than x² + 9. So clearly x < sqrt(x² + 9) < 1 + sqrt(x² + 9) for all x.
only real numbers can be compared.
there's still no solutions
@@maxvangulik1988 That x² + 9 is real (and indeed positive) is implicit in the use of 'sqrt´, since sqrt is a multivalued function on C and the convention used for real numbers that it denotes the non-negative square root cannot apply to complex numbers in general (or even to negative reals). It would be ambiguous notation without a determinate sense. So x² is also real. You are correct that this by itself leaves open the possibility that x is purely imaginary, but that would be obviously incompatible with x = 1 + sqrt(x^2 + 9), since the RHS is real (and indeed >= 4).
@@maxvangulik1988 Indeed, and my remark was intended as a simpler way of seeing this.
@@russellsharpe288 Well, there is something called the _principal_ square root of a complex number. The principal square root √z of a complex number z = r·exp(φi) with r ≥ 0, −π < φ ≤ π is _defined_ as √z = √r·exp(½φi) and, by convention, for any complex number z (which includes real numbers), the notation √z is commonly (but not universally) understood to refer to this uniquely defined principal value, which has a nonnegative real part. See the Wikipedia article on square roots. Wolfram Alpha also follows this convention. For example, if you enter √(3 + 4i) or sqrt(3 + 4i) then the (unique) answer is 2 + i, _not_ −2 − i.
You can check for a complex solution with x=a+bi:
(a-1+bi)^2=(a+bi)^2+9
a^2-2a+1 +2(a-1)bi -b^2 = a^2 +2abi -b^2 +9
a lot cancels out
-2a+1 -2bi = 9
a + bi = -4
Great work. Your answer a +bi =-4 is essentially the same answer he got -4,which doesn’t satisfy the original equation. If you compare real and imaginary parts a=-4 and b=0. This means there are no complex solutions either.
If we had x = 1 - sqrt(x^2 + 9) instead of the original problem, then we could have said x = -4 since -4 would have indeed equaled 1 - 5. The square-root radical means that only the principal square root can be considered. The principal square root is the one whose real part is positive in value. Therefore, the square root with the square-root radical or with the exponent "0.5" has to be 5 and not -5.
Generally speaking, the principal nth root of a number that is real, imaginary, or complex is the one that has the largest positive real part a if the heading is other then 180 degrees (negative real part and no imaginary part) and the one that has the largest positive real part a and has the positive imaginary part bi if the heading is 180 degrees. The angle or heading of the value (arctan of b/a) is considered to be in the range -180 degrees < angle
Also, it is a good idea to ask your teacher if you get in a situation in which you are using the 3th root or higher (4th, 5th, etc.) if he or she assumes the principal root or the root with the same sign as the number that you are taking the root of involving a situation similar to this one or if you are simply taking the root of a number under the radical. For example, the root normally accepted as the cube root of -8 is -2, but the principal cube root of -8 is actually 2*cos (60 degrees) + 2*sin(60 degrees) or 1 + sqrt(3)i.
So we just disown the negative Squareroot?
😁
The square root symbol means positive square root, so there is no negative square root unless we put a negative sign or a plus/minus sign in front of it.
The same is true with higher roots of a positive real. For cube roots, you need to put a w or w^2 in front of a cube root radical in order to have a root other than that real root as some call it. For square roots, w = -1. For cube roots, w = -0.5 + (sqrt(3)/2)i and w^2 = -0.5 - (sqrt(3)/2)i. For nth roots (n > 2), w = cos (360/n) + (sin (360/n))i, w^2 = cos (360*2/n) + (sin (360*2/n))i ... w^(n-1) = cos(360*(n-1)/n) + (sin(360*(n-1)/n))i. For nth roots (n > 1), w^n = 1.
I made x a complex number and solved the equation. Likely made a mistake, but I got Xr = -4 and Xi = 0. Hence, same result.
Yep. Me too. I guess there is no solution.
Yes, anyone reading this that doesn't see, substitute x=a+bi:
(a-1+bi)^2=(a+bi)^2+9
a^2-2a+1 +2(a-1)bi -b^2 = a^2 +2abi -b^2 +9
a lot cancels out
-2a+1 -2bi = 9
a + bi = -4
@teambellavsteamalice But -4 is extraneous so is not the right solution.
Sir i haven't solve a math problem 😔. Con you help me to solve it.
How can i knock you in online?
But x = -1 + sqrt(x²+9) HAS the solution x = 4.
That’s right
The problem here is that the radical symbol itself means "positive square root of."
There's nothing magical here. In any complex system, you can always array the symbols in such a way as to create nonsense.
Can you solve y+2=y This should be the new video
You can't. By subtracting y by both sides, you get 2 = 0, which is not true.
Exactly. No solutions other than infinity 😁
@@Manisphesto Seems fine in Z/2Z
@@SyberMath you're the best SyberMath. I know i can rely on you
@@SyberMathbut if you IGNORE ARBITRARU CONVENTION then the square root can be negative, so why don't you acknowledge that and say x CAN equal negative 4 also??
x = 4
Hi friend. Extraneous root. So basic. (x-1)^2=[-sqrt(x^2+9)]^2 ? I guess l said the answer. Good luck friend!!
hey im bhargav 😁😁
Тривиальное уравнение, 😂.
I think you can just plug in exp(it) = -4, then solve. Multiply both sides by exp([2n+1]𝜋i)=-1 to get exp([2n+1]𝜋i+it)= 4. Then take natural logs: [2n+1]𝜋i+it = ln(4), then multiply both sides by -i and solve for t = -(2n+1)𝜋-ln(4)i, then solve for x=exp(it) = exp(ln(4)-(2n+1)𝜋i) = -4. Oops! ;-) But seriously, is there no way to get an answer? Maybe not. Here's a simpler case:
√x = -3
You can't get a solution because the left side is positive and the right side is negative. You can't get a complex solution because √x can't have an imaginary component.
There is no solution.
No sol
If anyone found complex sol reply me pls.
If my math's right, I think the complex solutions are generated by: a+bi+4
@@MisterPenguin42 thx !
If you did an x = (a + ib) substitution you still have to introduce his term of an "extraneous" a = -4 solution. But you manipulate to: -2a - 8 = 2bi ... But a + ib was to assume and be real coefficients. This implies 2bi = 0 and -2a - 8 = 0 if 2bi has to equal 0. So again Z = -4 (no matter how many infinite angles Z = 4 at angles π + 2nπ) is the only Complex Plane solution to Z = 1 + √(Z^2 + 9) or -4 = 1 + √((-4)^2 + 9)!!
@lawrencejelsma8118 I'm sorry I did not understand your long response. Could you just straight up clearly respond in Yes or No as to whether you believe there is a Complex solution or not. If your answer is yes, then feel free to tell us what that complex root/roots are and how you got them?
@@trojanleo123 ... Yes or no ? Please go to school kid and learn to read ... I'm not going to respond to your dumb not able to do sentence by sentence reading for you. Math is not a yes or no answer!
There are no complex solutions either.
Square as is.
x^2=1+x^2+9+2sqrt(x^2+9)
sqrt(x^2+9) + 5 =0
If x is real LHS>0 so no real solutions. If x is complex LHS will generate a non zero imaginary part. On RHS the imaginary part is zero. So there are no complex solutions either.
No solutions. It's obvious.
If x=0, rhs is 4
-2x+1=9...x=-4...no solution
√25 = ±5
No, that radical symbol means "positive square root".
If my math's right, I think the complex solutions are generated by: a+bi+4
a + bi - 1 = sqrt((a+bi)^2+9)
a^2 + 2abi - 2a - b^2 - 2bi + 1 = a^2 - b^2 + 9 + 2abi
-2 - 2bi - 8 = 0
-2(a + bi + 4) = 0
a + bi + 4 = 0
So a = -4 and b = 0.
Look. -4 should be a correct solution.
-4 = 1 + sqrt(25)
But sqrt(25) = 5 or -5
-4 = 1 - 5
Sq root of 25 is strictly positive 5 , NOT + or - 5
sqrt(25) = 5. Not -5.