Solving x^(2/3)=64

How do we solve x^(2/3)=64? In this algebra tutorial, we will see how to solve an equation with a rational exponent. We can either do it from the exponent way (but we may make careless mistakes) or change it to the radical form first then solve. This is the type of equations you will encounter in Algebra 2 and Precalculus!

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#math #algebra #mathbasics

[AMVirtual-bxdi]

Average BPRP's video: 2 + 2 = x, x ≠ 4 😂

[kevinmadden]

Let the cube root of x equal u. Therefore u squared equals 64 and u equals 8 and -8. Therefore x equals (+ and -8) cubed. Therefore x equals 512 and -512.

[ChavoMysterio]

x^⅔=64
³√(x²)=64
x²=64³
x²=(8³)²
x²=512²
|x|=512
x=±512 ❤❤

[javierferrandizlarramona]

Hola. Sería mejor hacer x^2-64^3=0. Y ahora (x+sqr(64^3))•(x-sqr(64^3))=0 creo. Y así puedes sacar las dos soluciones. Poner el +- delante de una raíz puede llevar a pensar que el número de soluciones de una raíz cuadrada son dos, cuando eso no es cierto. De todas formas, gracias por tus ejercicios y por enseñarnos tanto.

[briancarlton]

What is your Expo budget, bprp? Love your videos, sir!

[samueladler]

I may write it as [x^(1÷3)]²—8²=0 diff of two squares.
[x^(1÷3)—8][x^(1÷3)+8]=0
Case 1
X^(1÷3)^3=8³
x=8³
Case II

X^(1÷3)^3=-8³

X=-8³

[discreaminant]

Gonna remind u guys that when the exponent is a non-integer real number, the base can only take positive real numbers

[DrFunkman]

”Can we see the mistake with this?”
Yes, but ignorance is bliss

[Kero-zctc]

If a graphed function can have vertical and horizontal asymptotes can it also have asymptotes with polynomial equations? What about sin x?

[cyruschang]

x^(2/3) = 64 = (+/- 8)^2
x^(1/3) = +/- 8
x = +/- 8^3 = +/- 512

[m.h.]

Instead of 2⁶, I would have just used 8², which cancels with the 3/2 to 8³, which is 512

[ArshKhan-qjkf]

How come the pinned comment was commented 10 hours before the video was uploaded?

[Limited_Light]

x^(1/3) = [plus or minus] 8, so x = [plus or minus] 8^3 = (2^3)^3 = 2^9 = 512. How about whether there are any complex solutions? (Ads still playing s I type this.)

[Trebatius]

Wait a minute! Why is x^(2/3) = (x^2)^(1/3), and not (x^(1/3))^2?

And where is the error then in the following calculation?
√x = 2
x^(1/2) = 2
x^(2/4) = 2
(x^2)^(1/4) = 2
x^2 = 16
x = ±4

[josephmercurio]

I've been solving these wrong all my life! 😮

[akonako]

That +/- sign always distracting me and cant focus on the board. :(

[DokkanPapaya]

The highest exponent is 2/3rd so there are 2/3 answers 😅

[major__kong]

You have to be even more careful. What if you take this approach: sqrt(x^3) = 64? Then x needs to be greater than 0. What rule says sqrt(x)^(1/3) is the correct representation?

[JimmyMatis-hy]

according to the fundamental theorem of algebra, shouldn't it have ⅔ of a root? 😋

[ronaldking]

Domain is all real numbers
x^(2/3) = 64 -> (x^(1/3))^2 = 64
Let y = x^(1/3) -> y^2 = 64
y^2 - 64 = 0 -> (y-8)(y+8) = 0
y = +/-8 ->x^1/3 = +/-8
x = (+/-8)^3 = +/-512
Domain is all real numbers and as such, both check out.