Q. 8.30 - Intermediate Algebra

Question

Simplify-

(a) $\sqrt{p^9}$

(b) $\sqrt[5]{y^8}$

(c) $\sqrt[6]{q^{13}}$

Step-by-Step Solution

Answer

The values are-

(a) $p^4\sqrt{p}$

(b) $y\sqrt[5]{y^3}$

(c)  $q^2 \sqrt[6]{q}$

Step 1. Explanation

(a) We will write the radical in form of product of two radicals with one being the largest perfect square factor of the number. Then we will simplify it to get the value. 

(b) We will write the radical in form of product of two radicals with one being the largest perfect fifth power factor of the number. Then we will simplify it to get the value.  

(c) We will write the radical in form of product of two radicals with one being the largest perfect sixth power factor of the number. Then we will simplify it to get the value.  

Step 2. Solving for part (a)

We have $\sqrt{p^9}$.

We have $p^8$ as the largest perfect square factor of $p^9$.

So this can be written as $\sqrt{p^8\times p}$.

Using the product rule, we get $\sqrt{p^8}\times \sqrt{p}$.

We know that ${\left(p^4\right)}^2=p^8$.

So using this, we get the value as 

$$\begin{gathered} =\sqrt{{\left(p^4\right)}^2}\times \sqrt{p} \\ =p^4\sqrt{p} \end{gathered}$$

Step 3. Solving for part (b)

We have $\sqrt[5]{y^8}$.

We have $y^5$ as the largest fifth power factor of $y^8$.

So this can be written as $\sqrt[5]{y^5\times y^3}$.

Using the product rule we get $\sqrt[5]{y^5}\times \sqrt[5]{y^3}$.

This gives us the value as $y\sqrt[5]{y^3}$.

Step 4. Solving for part (c)

We have $\sqrt[6]{q^{13}}$.

We have $q^{12}$ as the largest perfect sixth power factor of $q^{13}$.

So this can be written as $\sqrt[6]{q^{12}\times q}$.

Using the product rule we get $\sqrt[6]{y^{12}}\times \sqrt[6]{q}$.

We know that $y^{12}=y^6\times y^6$.

So using it, we get $\sqrt[6]{y^6}\times \sqrt[6]{y^6}\times \sqrt[6]{q}$.

This gives us the value as $q^2 \sqrt[6]{q}$.