2x^{5/3}+64=0

## 2x5/3 +64=0

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# 3 Answers

2x^{5/3} + 64 = 0

Subtract 64 from both sides of the equation:

2x^{5/3}^{ }= -64

Divide both sides of the equation by 2:

x^{5/3} = -32

Cube both sides of the equation:

(x^{5/3})^{3} = (-32)^{3}

x^{5} = -32768

Take both sides of the equation to the power of 1/5:

(x^{5})^{1/5} = (-32768)^{1/5}

x = -8

This problem may have been written incorrectly, as it only has complex number solutions (not part of algebra 1).

The exact (complex) solutions are:

x = 8(-1)^{3/5}

x = -8(-1)^{2/5}

Note: if you change the plus sign in** 2x ^{5/3 }+ 64=0** to a minus sign:

** 2x ^{5/3 }- 64=0**

then follow Tamara's working, you'll get x = 8

2x^(5/3)+64=0 Solution 2x5/3+64=0 x^(5/3) = -64/2 x^(5/3) = -32 x^5 = -32^3 x^5 = -32768 x = (-32768)^(1/5) x = (32768)^(1/5) i, where i = (-1)^(1/2) x=8i =========================================================================== Step by Step solution
Our goald is to write an equation with a form x = ? 1st step, sume -64 both places of the equation. That is because you do not want the +64 at the left side of the equation. 2x^(5/3) = -64 2nd step, divide by 1/2 all the equation. That is because you do not
want 2x, you want x only x^(5/3) = -64/2 = -32 x^(5/3) = -64/2, then x^(5/3) = -32 3rd step, raises the whole equation to 3/5 exponent [x^(5/3) = -32]^(3/5) [x^(5/3)]^(3/5) = [-34]^(3/5) 4st stepUse the exponent property (a^n)^m = a^(nm) x^(5*3/3*5)= [-34]^(3/5)
x = [-34]^(3/5)= R(5) [-34^3], wher R()[?]= fifth root of ? note: R(5)[34^3*(-1)^3]=R(5)[39304*-1]=R(5)[39304]i x=8i

# Comments

8i also doesn't appear to be a valid solution to the original equation

## Comments

-8 is not a valid solution when you plug it back into the original equation

How come? (-8)^(5/3) = -32

2(-32) + 64 = 0

I agree with Tamara.

There are no complex number solutions when there is an odd index (i.e. raising to the 1/5th is taking the 5th root) Taking an odd root of a negative number is just negative. Taking an even root of a negative number yields complex solutions.

Greg, there still can be complex solutions when taking an odd root, there just happens to be a real root as well (the principal roots are complex): http://mathworld.wolfram.com/CubeRoot.html

eg. for x^5 = -32768 there are five roots: http://www.wolframalpha.com/input/?i=x%5E5+%3D+-32768

If you scroll down to the plot of the roots in the complex plane, you can see that there's one real root and four imaginary roots.

The number of roots will match the exponent - an exponent of n has n roots (whether n is even or odd, and whether the base is negative or positive), eg x^32 = 1 has 32 roots, two of them real (±1):

-8 is actually a solution to the equation, so I was wrong there.

For x^32 = 1, you can see the plot of the roots: http://www.wolframalpha.com/input/?i=x%5E32+%3D+1+;

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