Q. 286 - Intermediate Algebra

Question

Explain why multiplying $\frac{7}{\sqrt[3]{x}}$by$\frac{\sqrt[3]{x}}{\sqrt[3]{x}}$

does not rationalize the denominator.

Step-by-Step Solution

Answer

In order to rationalize the denominator, multiply numerator and denominator by an expression such that denominator results into no radicals. In the given expression $\frac{7}{\sqrt[3]{x}}$,the denominator $\sqrt[3]{x}$ has only one power of $x$.Hence, multiply numerator and denominator by $\sqrt[3]{x^2}$ to make the perfect cube. Thus, multiplying the given expression $\frac{7}{\sqrt[3]{x}}$ by $\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}$ results in rationalization.

Step 1 The given expression is,

$\frac{7}{\sqrt[3]{x}}$.

The objective is, we need to give the reason for multiplying the expression $\frac{7}{\sqrt[3]{x}}$ by $\frac{\sqrt[3]{x}}{\sqrt[3]{x}}$ does not result in rationalization.

Step 2 Reason for multiplying the given expression by x 3 x 3 does not result in rationalization :

In order to rationalize the denominator, multiply numerator and denominator by an expression such that denominator results into no radicals. In the given expression $\frac{7}{\sqrt[3]{x}}$, the denominator $\sqrt[3]{x}$ has only one power of $x$. Hence, multiply numerator and denominator by $\sqrt[3]{x^2}$ to make the perfect cube.

Thus, multiplying the given expression $\frac{7}{\sqrt[3]{x}}$ by $\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}$ results in rationalization.

Therefore, the reason for multiplying the expression $\frac{7}{\sqrt[3]{x}}$ by $\frac{\sqrt[3]{x}}{\sqrt[3]{x}}$ does not result in rationalization is explained.