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# Simplifying square roots with absolute value

The absolute value $|a|$ of a real number $a$ may be thought of as its distance from zero (From geometry point of view).

The most important definition of absolute value together with square root.

$$\sqrt{a^2}=|a|=\begin{cases} a, & \text{if} \ \ \ a \ge 0 \\ -a, & \text{if} \ \ \ a<0 \end{cases}$$

## Solved examples

Simplify the expression $\sqrt{(2-\sqrt{3})^2}$

$$\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}$$

Simplify the expression $\sqrt{(4-\sqrt{17})^2}$

$$\sqrt{(4-\sqrt{17})^2}=|4-\sqrt{17}|=-(4-\sqrt{17})=-4+\sqrt{17}$$

Simplify the expression $\sqrt{(3-\sqrt{5})^2}-\sqrt{(\sqrt{5}-4)^2}$

$$\sqrt{(3-\sqrt{5})^2}-\sqrt{(\sqrt{5}-4)^2}=|3-\sqrt{5}|-|\sqrt{5}-4|=|3-\sqrt{5}|-|\sqrt{5}-4|=3-\sqrt{5}+\sqrt{5}-4=-1$$

Simplify the expression $\sqrt{(x-3)^2}-\sqrt{(3-x)^2}$, if $x>3$

$$\sqrt{(x-3)^2}-\sqrt{(3-x)^2}=|x-3|-|3-x|=x-3+3-x=0$$

Simplify the expression $\sqrt{x^2+\frac{11}{25}x^2}-\sqrt{\frac{1}{25}x^2}$, if $x<0$

$$\sqrt{x^2+\frac{11}{25}x^2}-\sqrt{\frac{1}{25}x^2}=\sqrt{\frac{36}{25}x^2}-\sqrt{\frac{1}{25}x^2}=\left |\frac{6}{5}x\right |-\left |\frac{1}{5}x\right |=-\frac{6}{5}x+\frac{1}{5}x=-x$$