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# Solving Quadratic Equations with Square Roots

## Solving Quadratic Equations with Square Roots 1.24 - Solution

a

We use inverse operations to isolate the $x$-terms on one side.

$\dfrac{x}{3}=\dfrac{5}{x}$
$x=\dfrac{15}{x}$
$x^2=15$
$x=\pm \sqrt{15}$
Both $x=\text{-} \sqrt{15}$ and $x=\sqrt{15}$ are solution to the equation.
b

Again, we will use inverse operations in order to get the $x$-terms isolated on one side.

$\dfrac{3x}{6}-\dfrac{8}{x}=0$
$\dfrac{3x}{6}=\dfrac{8}{x}$
$\dfrac{x}{2}=\dfrac{8}{x}$
$x=\dfrac{16}{x}$
$x^2=16$
$x=\pm \sqrt{16}$
$x=\pm 4$

Both $x=\text{-} 4,$ and $x=4$ are solutions to the equation.

c

The left-hand side's least common denominator is $6x.$ We will use that to rewrite the LHS as a single fraction. After that we will solve the equation using inverse operations.

$\dfrac{5}{2x}-\dfrac{1}{3x}=\dfrac{x}{6}$
$\dfrac{15}{6x}-\dfrac{1}{3x}=\dfrac{x}{6}$
$\dfrac{15}{6x}-\dfrac{2}{6x}=\dfrac{x}{6}$
$\dfrac{13}{6x}=\dfrac{x}{6}$
$13=\dfrac{6x^2}{6}$
$13=x^2$
$x^2=13$
$x=\pm\sqrt{13}$

$x=\text{-} \sqrt{13}$ and $x=\sqrt{13}$ are solutions to the equation.