Chapter R.7, Problem 1YT

To determine

To simplify: The expression $\sqrt{28x^9{y}^5}$.

Answer to Problem 1YT

The simplified form of the expression $\sqrt{28x^9{y}^5}$ is $2x^4{y}^2\sqrt{7xy}$.

Explanation of Solution

Given:

The expression $\sqrt{28x^9{y}^5}$.

Definition used:

Radicals: If n is an even natural number and $a>0$, or n is an odd natural number, then $a^{1/n}=\sqrt[n]{a}$.

Property used:

1. For all real numbers a and natural number n such that $\sqrt[n]{a}$ is real number, ${\left(\sqrt[n]{a}\right)}^n=a$.

3. For all real numbers a and b and natural numbers m and n such that $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers, $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{ab}$.

Calculation:

Simplify the terms,

$\begin{array}{c}\sqrt{28x^9{y}^5}=\sqrt{\left(4\cdot 7\right)\left(x^8\cdot x\right)\left(y^4\cdot y\right)}\left[\text{By Property 3}\right]\\ =\sqrt{\left(4\right)\left(x^8\right)\left(y^4\right)\cdot \left(7\cdot x\cdot y\right)}\\ =\sqrt{\left(4\right)\left(x^8\right)\left(y^4\right)\cdot \left(7xy\right)}\\ =\sqrt{\left(4\right)\left(x^8\right)\left(y^4\right)}\cdot \sqrt{\left(7xy\right)}\end{array}$

By the definition of radicals,

$\sqrt{28x^9{y}^5}={\left(\left(4\right)\left(x^8\right)\left(y^4\right)\right)}^{1/2}\cdot \sqrt{7xy}$

Further simplify the terms of the expression as follows,

$\begin{array}{c}\sqrt{28x^9{y}^5}={\left(4\right)}^{1/2}{\left(x^8\right)}^{1/2}{\left(y^4\right)}^{1/2}\cdot \sqrt{7xy}\left[\text{By Property 1}\right]\\ ={\left(2^2\right)}^{1/2}{\left(x^8\right)}^{1/2}{\left(y^4\right)}^{1/2}\cdot \sqrt{7xy}\\ =\left(2^{\left(2\right)}{}^{\left(1/2\right)}\right)\left(x^{\left(8\right)\left(1/2\right)}\right)\left(y^{\left(4\right)\left(1/2\right)}\right)\cdot \sqrt{7xy}\\ =2x^4{y}^2\sqrt{7xy}\end{array}$

Therefore, the simplified form of the expression $\sqrt{28x^9{y}^5}$ is $2x^4{y}^2\sqrt{7xy}$.