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To begin our study of the process of simplifying a square root expression, we must note three facts: one fact concerning perfect squares and two concerning properties of square roots.
Although we will not make a detailed study of irrational numbers, we will make the following observation:
Any indicated square root whose radicand is not a perfect square is an irrational number.
The numbers $\sqrt{6},\text{\hspace{0.17em}}\sqrt{15},$ and $\sqrt{\frac{3}{4}}$ are each irrational since each radicand $\left(6,\text{\hspace{0.17em}}15,\text{\hspace{0.17em}}\frac{3}{4}\right)$ is not a perfect square.
Notice that
$\sqrt{9\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}4}=\sqrt{36}=6$ and
$\sqrt{9}\text{\hspace{0.17em}}\sqrt{4}=3\text{\hspace{0.17em}}\xb72\text{\hspace{0.17em}}=6$
Since both
$\sqrt{9\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}4}$ and
$\sqrt{9}\text{\hspace{0.17em}}\sqrt{4}$ equal 6, it must be that
$\sqrt{9\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}4}=\sqrt{9}\sqrt{4}$
We can suggest a similar rule for quotients. Notice that
$\sqrt{\frac{36}{4}}=\sqrt{9}=3$ and
$\frac{\sqrt{36}}{\sqrt{4}}=\frac{6}{2}=3$
Since both
$\frac{36}{4}$ and
$\frac{\sqrt{36}}{\sqrt{4}}$ equal 3, it must be that
$\sqrt{\frac{36}{4}}=\frac{\sqrt{36}}{\sqrt{4}}$
CAUTION
It is extremely important to remember that
$\begin{array}{lllll}\sqrt{x+y}\ne \sqrt{x}+\sqrt{y}\hfill & \hfill & \text{or}\hfill & \hfill & \sqrt{x-y}\ne \sqrt{x}-\sqrt{y}\hfill \end{array}$
For example, notice that $\sqrt{16+9}=\sqrt{25}=5,$ but $\sqrt{16}+\sqrt{9}=4+3=7.$
We shall study the process of simplifying a square root expression by distinguishing between two types of square roots: square roots not involving a fraction and square roots involving a fraction.
A square root that does not involve fractions is in simplified form if there are no perfect square in the radicand.
The square roots $\sqrt{x,}\sqrt{ab},\sqrt{5mn,}\sqrt{2\left(a+5\right)}$ are in simplified form since none of the radicands contains a perfect square.
The square roots $\sqrt{{x}^{2},}\sqrt{{a}^{3}}=\sqrt{{a}^{2}a}$ are not in simplified form since each radicand contains a perfect square.
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