1.13: Simplifying Square Roots

Finding a square root of a number is the inverse operation of squaring that number. Remember, the square of a number is that number times itself. For example,

\[5^{2}=5 \cdot 5=25 \nonumber\]

and

\[(-5)^{2}=(-5) \cdot(-5)=25. \nonumber\]

The square root of a number \(n,\) written as \(\sqrt{n},\) is the positive number that gives \(n\) when multiplied by itself. For example, \(\sqrt{25}=5\) and not -5 because 5 is the positive number that multiplied by itself gives 25.

The perfect squares are the squares of whole numbers:

• \(1=1^{2}\)
• \(4=2^{2}\)
• \(9=3^{2}\)
• \(16=4^{2}\)
• \(25=5^{2}\)
• \(36=6^{2}\)
• \(49=7^{2}\)
• \(64=8^{2}\)
• \(81=9^{2}\)
• \(100=10^{2}, \ldots\)

and finding their square roots is straightforward. So

\[\begin{align*} \sqrt{16} &=\sqrt{4^{2}} \\[4pt] &=4 \\ \sqrt{100} &= \sqrt{10^{2}} \\[4pt] &=10. \end{align*}\]

What about \(\sqrt{50} ?\) Can you think of a number you multiply by itself and the answer is \(50 ?\) The only thing we can do is simplify the square root. We say that a square root is simplified if it has no perfect square factors.

So, to simplify \(\sqrt{50}\) we first write 50 into its factor and look for perfect squares.

\[50=25 \cdot 2=5^{2} \cdot 2 . \nonumber\]

Then,

\[\begin{align*} \sqrt{50} &=\sqrt{25 \cdot 2} \\[4pt] &=\sqrt{5^{2}} \cdot \sqrt{2} \\[4pt] &=5 \cdot \sqrt{2}. \end{align*}\]

The justification for separating \(\sqrt{5^{2}}\) and \(\sqrt{2}\) is the fact that the square root of a product is equal to the product of the square root of each factor:

Product Rule for Radicals

\[\sqrt{a \cdot b}=\sqrt{a} \cdot \sqrt{b}\]

Example 11.1

Simplify each of the following radical expressions:

1. \(\begin{align*} \sqrt{24} &=\sqrt{4 \cdot 6}=\sqrt{2^{2} \cdot 6} \\ &=\sqrt{2^{2}} \cdot \sqrt{6}=2 \cdot \sqrt{6} \end{align*}\)
2. \(\sqrt{108}=\sqrt{36 \cdot 3}=\sqrt{6^{2} \cdot 3}=\sqrt{6^{2}} \cdot \sqrt{3}=6 \cdot \sqrt{3}\)
3. \(\begin{align*} 2 \cdot \sqrt{80} &=2 \cdot \sqrt{16 \cdot 5}=2 \cdot \sqrt{4^{2} \cdot 5} \\ &=2 \cdot \sqrt{4^{2}} \cdot \sqrt{5}=2 \cdot 4 \cdot \sqrt{5} \\ &=8 \cdot \sqrt{5} \end{align*}\)

How to combine “like” square roots

We can combine “like” square roots the same way we combined “like terms” in chapter 8.

Like Square Roots

Two (or more) square roots are "like" if they have the same quantity under the root.

Note: Always simplify the square root if possible before identifying "like" roots.

Example 11.2

Like square roots:

1. \(\sqrt{3}\) and \(-6 \sqrt{3}\)
2. \(2 \sqrt{5}\) and \(-4 \sqrt{5}\)
3. \(\sqrt{7}\) and \(\sqrt{28},\) because \(\sqrt{28}=\sqrt{4 \cdot 7}=\sqrt{4} \cdot \sqrt{7}=2 \sqrt{7}\)
4. \(\sqrt{90}\) and \(\sqrt{250}\) because \(\sqrt{90}=\sqrt{9 \cdot 10}=\sqrt{9} \cdot \sqrt{10}=3 \sqrt{10}\) and \(\sqrt{250}=\sqrt{25 \cdot 10}=\sqrt{25} \cdot \sqrt{10}=5 \sqrt{10}\)

To add or to subtract radicals, we need to first simplify radicals, then combine like radicals.

Example 11.3

Add or Subtract radicals:

1. \(\sqrt{160}+\sqrt{490}=\sqrt{16 \cdot 10}+\sqrt{49 \cdot 10}=\sqrt{16} \cdot \sqrt{10}+\sqrt{49} \cdot \sqrt{10}=4 \cdot \sqrt{10}+7 \cdot \sqrt{10}=(4+7) \sqrt{10}=11 \sqrt{10}\)
2. \(2 \sqrt{27}-5 \sqrt{3}=2 \sqrt{9 \cdot 3}-5 \sqrt{3}=2 \sqrt{9} \cdot \sqrt{3}-5 \sqrt{3}=2 \cdot 3 \sqrt{3}-5 \sqrt{3}=6 \sqrt{3}-5 \sqrt{3}=(6-5) \sqrt{3}=1 \sqrt{3}=\sqrt{3}\)
3. \(4 \sqrt{18}-7 \sqrt{8}-3 \sqrt{1}=4 \sqrt{9} \sqrt{2}-7 \sqrt{4} \sqrt{2}-3 \cdot 1=4 \cdot 3 \sqrt{2}-7 \cdot 2 \sqrt{2}-3=12 \sqrt{2}-14 \sqrt{2}-3=(12-14) \sqrt{2}-3=-2 \sqrt{2}-3\)

Note: Only “like” roots can be combined.

How do we simplify non-numerical radicals?

Similar to numbers, variables inside the square root that are squared (raised to the power 2 ) can be simplified. So, \(\sqrt{x^{2}}=x\) in the same way that \(\sqrt{8^{2}}=8\). So, we need to find as many multiples of variables that are squared:

\[\begin{align*} \sqrt{x^{8}} &=\sqrt{x^{2} \cdot x^{2} \cdot x^{2} \cdot x^{2}} \\[4pt] &=\sqrt{x^{2}} \cdot \sqrt{x^{2}} \cdot \sqrt{x^{2}} \cdot \sqrt{x^{2}} \\[4pt] &=x \cdot x \cdot x \cdot x \\[4pt] &=x^{4} \\[4pt] \sqrt{x^{5}} &=\sqrt{x^{2} \cdot x^{2} \cdot x} \\[4pt] &=\sqrt{x^{2}} \cdot \sqrt{x^{2}} \cdot \sqrt{x} \\[4pt] &=x \cdot x \cdot \sqrt{x} \\[4pt] =x^{2} \sqrt{x}\end{align*}\]

If you have a number and a variable inside the square root (or more than one variable), you work with each one separately. For example:

\[\sqrt{50 x^{8}}=\sqrt{50} \cdot \sqrt{x^{8}}=5 \sqrt{2} \cdot x^{4}=5 x^{4} \sqrt{2}\nonumber\]

Example 11.4

Simplify each of the following radical expressions:

a) \(\begin{align*}
\sqrt{y^{4} x^{8}} &=\sqrt{y^{4}} \cdot \sqrt{x^{8}}=\sqrt{y^{2} \cdot y^{2}} \cdot \sqrt{x^{8}} \\
&=\sqrt{y^{2}} \cdot \sqrt{y^{2}} \cdot \sqrt{x^{8}}=y \cdot y \cdot \sqrt{x^{8}}=y^{2} \cdot \sqrt{x^{8}} \\
&=y^{2} \cdot x^{4} \\
&=y^{2} x^{4}
\end{align*}\)

b) \(\begin{align*}
\sqrt{200 \cdot m^{4}} &=\sqrt{200} \cdot \sqrt{m^{4}}=\sqrt{100 \cdot 2} \cdot \sqrt{m^{4}} \\
&=\sqrt{10^{2} \cdot 2} \cdot \sqrt{m^{4}}=\sqrt{10^{2}} \cdot \sqrt{2} \cdot \sqrt{m^{4}}=10 \cdot \sqrt{2} \cdot m^{2} \\
&=10 \cdot m^{2} \cdot \sqrt{2}=10 m^{2} \sqrt{2}
\end{align*}\)

c) \(\begin{align*}
m^{3} \cdot \sqrt{200 \cdot m^{4}} &=m^{3} \cdot \sqrt{200} \cdot \sqrt{m^{4}}=m^{3} \cdot \sqrt{100 \cdot 2} \cdot \sqrt{m^{4}} \\
&=m^{3} \cdot \sqrt{10^{2} \cdot 2} \cdot \sqrt{m^{4}}=m^{3} \cdot \sqrt{10^{2}} \cdot \sqrt{2} \cdot \sqrt{m^{4}} \\
&=m^{3} \cdot 10 \cdot \sqrt{2} \cdot m^{2}=10 m^{5} \cdot \sqrt{2} \\
&=10 m^{5} \sqrt{2}
\end{align*}\)

d) \(\begin{align*}
2 \sqrt{63 x^{3}} &=2 \cdot \sqrt{63} \cdot \sqrt{x^{3}}=2 \cdot \sqrt{9 \cdot 7} \cdot \sqrt{x^{3}} \\
&=2 \cdot \sqrt{3^{2} \cdot 7} \cdot \sqrt{x^{2} \cdot x}=2 \cdot \sqrt{3^{2}} \cdot \sqrt{7} \cdot \sqrt{x^{2}} \cdot \sqrt{x} \\
&=2 \cdot 3 \sqrt{7} \cdot x \cdot \sqrt{x}=6 x \cdot \sqrt{7 x}=6 x \sqrt{7 x}
\end{align*}\)

Similar to the product rule, the quotient rule allows us to separate square roots as follows:

Quotient Rule for Radicals

\(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\)

And notice that both rules can be read from right to left as follows:

\[\sqrt{a} \cdot \sqrt{b}=\sqrt{a \cdot b}\nonumber\]

\[\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\nonumber\]

This provides us with the necessary tools to combine and simplify square roots, when the operation is multiplication or division only.

Example 11.5

Simplify completely:

1. \(\begin{align*} \frac{\sqrt{15} \sqrt{70}}{\sqrt{5}} &=\frac{\sqrt{15 \cdot 70}}{\sqrt{5}} \\ &=\sqrt{\frac{15 \cdot 70}{5}}=\sqrt{\frac{5 \cdot 3 \cdot 7 \cdot 5 \cdot 2}{5}} \\ &=\sqrt{5 \cdot 3 \cdot 2 \cdot 7}=\sqrt{210} \end{align*}\)
2. \(\begin{align*} -x \sqrt{12 y^{3}} \cdot 3 y^{2} \sqrt{15 x} &=-3 x y^{2} \cdot \sqrt{12 y^{3} \cdot 15 x}=-3 x y^{2} \cdot \sqrt{12 \cdot 15 x y^{3}} \\ &=-3 x y^{2} \cdot \sqrt{4 \cdot 3 \cdot 3 \cdot 5 x y^{3}} \\ &=-3 x y^{2} \cdot \sqrt{4 \cdot 3^{2} \cdot 5 x y \cdot y^{2}} \\ &=-3 x y^{2} \cdot \sqrt{4} \cdot \sqrt{3^{2}} \cdot \sqrt{5 x y} \cdot \sqrt{y^{2}} \\ &=-3 x y^{2} \cdot 2 \cdot 3 \cdot \sqrt{5 x y} \cdot y \\ &=-18 x y^{3} \cdot \sqrt{5 x y} \end{align*}\)

Exit Problem

Simplify: \(5 \sqrt{24}-2 \sqrt{54}-3 \sqrt{16}\)