If x is a negative real number such that \sqrt{(-x)^3 \cdot x^2} = 32, what is the value o… — GRE General Test

If $x$ is a negative real number such that $\sqrt{(-x)^3 \cdot x^2} = 32$ , what is the value of $x$ ?

-4

Simplifying the expression inside the radical gives

(-x)^3 \cdot x^2 = -x^5

. Setting

\sqrt{-x^5} = 32

and squaring both sides gives

-x^5 = 32^2 = 1024

, which means

x^5 = -1024

. The fifth root of

-1024

-4

. Since

-4

is a negative real number, it satisfies all conditions of the problem.

Simplify the expression under the square root

Since

(-x)^3 = -x^3

, we have

(-x)^3 \cdot x^2 = (-x^3) \cdot x^2 = -x^5

Applying exponent addition rules

x^a \cdot x^b = x^{a+b}

and odd power rules for negative quantities.

Square both sides of the equation to eliminate the square root

-x^5 = 32^2 = 1024

Squaring both sides of

\sqrt{-x^5} = 32

isolates the radicand.

Solve for x

x^5 = -1024 \implies x = \sqrt[5]{-1024} = -4

Taking the 5th root of

-1024 = (-2)^{10} = (-4)^5

yields

x = -4

, which satisfies the given condition

x < 0

Simplifying expressions with powers and square roots involving negative variables.

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