Q. 8.28 - Intermediate Algebra

Question

Simplify-

(a) $\sqrt{432}$

(b) $\sqrt[3]{625}$

(c) $\sqrt[4]{729}$

Step-by-Step Solution

Answer

The values are-

(a) $12\sqrt{3}$

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

(c) $3\sqrt{3}$

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 as the product using the largest perfect cube factor. Then simplifying the expression will give us the value.

(c)  We will write the radical as the product using the largest perfect fourth power factor. Then simplifying the expression will give us the value. 

Step 2. Solving part (a).

We have $\sqrt{432}$.

We have $144$ as the largest perfect square factor of $432$ as $144\times 3=432$ and
${\left(12\right)}^2=144$.

So we can write this as $\sqrt{144\times 3}$.

Writing it as product of two radicals gives us $\sqrt{144}\times \sqrt{3}$.

Solving it gives us the value as $12\sqrt{3}$.

Step 3.Solving for part (b)

We have $\sqrt[3]{625}$.

We have $125$ as the largest perfect cube factor of $625$ as $125\times 5=625$ and
${\left(5\right)}^3=125$.

So we can write this as $\sqrt[3]{125\times 5}$.

Writing it as product of two radicals gives us $\sqrt[3]{125}\times \sqrt[3]{5}$.

Solving it gives us the value as $5\sqrt[3]{5}$.

Step 4. Solving for part (c).

We have $\sqrt[4]{729}$.

We have $81$ as the largest perfect fourth power factor of $729$ as $81\times 9=729$ and
${\left(3\right)}^4=81$.

So we can write this as $\sqrt[4]{81\times 9}$.

Writing it as product of two radicals gives us $\sqrt[4]{81}\times \sqrt[4]{9}$.

This can be written as $\sqrt[4]{{\left(3\right)}^4}\times \sqrt[4]{{\left(\sqrt{3}\right)}^4}$.

So, solving the equation gives us the value as $3\sqrt{3}$.