Exercises: Terms with Square Roots

When solving quadratic equations, one obtains expressions with square roots, like the following two. Simplify them.

$x_{1/2}=\frac{-5\pm\sqrt{5^2+4\cdot\sqrt{7}\cdot2\sqrt{7}}}{2\sqrt{7}}$

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$\displaystyle x_{1/2}$	$=$	$\displaystyle \frac{-14\pm \sqrt{14^2-4\cdot 8}}{2}$
	↓	Conclude everything under the root.
	$=$	$\displaystyle \frac{-14\pm \sqrt{164}}{2}$
	↓	Pull 2 out of the root.
	$=$	$\displaystyle \frac{-14\pm \sqrt{4\cdot 41}}{2}=\frac{-14\pm 2\sqrt{41}}{2}$
	↓	Shorten the fractions.
	$=$	$\displaystyle -7\pm \sqrt{41}$

$\displaystyle x_{1/2}$	$=$	$\displaystyle \frac{-5\pm \sqrt{5^2+4\cdot \sqrt{7}\cdot 2\sqrt{7}}}{2\sqrt{7}}$
	↓	Conclude everything under the root.
$\displaystyle x_{1/2}$	$=$	$\displaystyle \frac{-5\pm \sqrt{81}}{2\sqrt{7}}=\frac{-5\pm 9}{2\sqrt{7}}$
$\displaystyle x_1$	$=$	$\displaystyle \frac{-5+9}{2\sqrt{7}}=\frac{4}{2\sqrt{7}}$
	↓	Shorten by 2
$\displaystyle x_1$	$=$	$\displaystyle \frac{2}{\sqrt{7}}$
	↓	Expand by $\sqrt{7}$ .
$\displaystyle x_1$	$=$	$\displaystyle \frac{2\cdot \sqrt{7}}{\sqrt{7}\cdot \sqrt{7}}=\frac{2\sqrt{7}}{7}$
$\displaystyle x_2$	$=$	$\displaystyle \frac{-5-9}{2\sqrt{7}}=\frac{-14}{2\sqrt{7}}$
	↓	Expand by $\sqrt{7}$ .
$\displaystyle x_2$	$=$	$\displaystyle \frac{-14\sqrt{7}}{2\sqrt{7}\sqrt{7}}=\frac{-14\sqrt{7}}{2\cdot 7}=\frac{-14\sqrt{7}}{14}$
	↓	Shorten by 14.
$\displaystyle x_2$	$=$	$\displaystyle -\sqrt{7}$