Q. 8.95 - Intermediate Algebra

Question

Simplify:

(a) $\frac{\sqrt{162x^{10}{y}^2}}{\sqrt{2x^6{y}^6}}$

(b) $\frac{\sqrt[3]{128x^2{y}^{-1}}}{\sqrt[3]{2x^{-1}{y}^2}}$

Step-by-Step Solution

Answer

The simplification of $\frac{\sqrt{162x^{10}{y}^2}}{\sqrt{2x^6{y}^6}}$ is $\frac{9x^2}{y^2}$

The simplification of $\frac{\sqrt[3]{128x^2{y}^{-1}}}{\sqrt[3]{2x^{-1}{y}^2}}$ is $\frac{4x}{y}$

Part (a) Step 1. Given

An expression $\frac{\sqrt{162x^{10}{y}^2}}{\sqrt{2x^6{y}^6}}$

To simplify the expression

Part (a) Step 2. Rewrite using the quotient property

Rewrite using the quotient property 

$\frac{\sqrt{162x^{10}{y}^2}}{\sqrt{2x^6{y}^6}}=\sqrt{\frac{162x^{10}{y}^2}{2x^6{y}^6}}$

Remove the common factors.

                     $=\sqrt{\frac{81x^4}{y^4}}$

Part (a) Step 3. Simplify the expression

$\frac{\sqrt{162x^{10}{y}^2}}{\sqrt{2x^6{y}^6}}=\sqrt{\frac{81x^4}{y^4}}$

                     $=\sqrt{\frac{9^2{\left(x^2\right)}^2}{{\left(y^2\right)}^2}}$

                      $=\frac{9x^2}{y^2}$

The simplification of the expression is $\frac{9x^2}{y^2}$

Part (b) Step 1. Given

An expression $\frac{\sqrt[3]{128x^2{y}^{-1}}}{\sqrt[3]{2x^{-1}{y}^2}}$

To simplify the expression 

Part (b) Step 2. Rewrite using the quotient property

Rewrite using the quotient property 

$\frac{\sqrt[3]{128x^2{y}^{-1}}}{\sqrt[3]{2x^{-1}{y}^2}}=\sqrt[3]{\frac{128x^2{y}^{-1}}{2x^{-1}{y}^2}}$

Remove the common factors.

                      $=\sqrt[3]{\frac{64x^3}{y^3}}$

Part (b) Step 3. Simplify the expression

$\frac{\sqrt[3]{128x^2{y}^{-1}}}{\sqrt[3]{2x^{-1}{y}^2}}=\sqrt[3]{\frac{64x^3}{y^3}}$

                      $=\sqrt[3]{\frac{4^3{x}^3}{y^3}}$

                      $=\frac{4x}{y}$

The simplification of the expression is $\frac{4x}{y}$