Solving Quadratic Equations with Square Roots

a

Add $40$ to both sides in order to isolate $4x^2.$ Then divide both sides by $4$ and, lastly, square root both sides of the equation.

4x^2-40=0

\text{LHS}+40=\text{RHS}+40

4x^2=40

\left.\text{LHS}\middle/4\right.=\left.\text{RHS}\middle/4\right.

x^2=10

\sqrt{\text{LHS}}=\sqrt{\text{RHS}}

x=\pm \sqrt{10}

Both $x=\text{-}\sqrt{10}$ and $x=\sqrt{10}$ are solutions to the equation.

b

We will begin by dividing by $9.$ Then we will square root both sides.

9x^2=\dfrac{4}{9}

\left.\text{LHS}\middle/9\right.=\left.\text{RHS}\middle/9\right.

x^2=\left.\dfrac{4}{9}\middle/9\right.

\left.\dfrac{a}{b}\middle/c\right.= \dfrac{a}{b\cdot c}

x^2=\dfrac{4}{81}

\sqrt{\text{LHS}}=\sqrt{\text{RHS}}

x=\pm \sqrt{\dfrac{4}{81}}

\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}

x=\pm \dfrac{\sqrt{4}}{\sqrt{81}}

SimpRootSimplify root

x=\pm \dfrac{2}{9}

Both $x=\text{-}\frac{2}{9}$ and $x=\frac{2}{9}$ are solutions to the equation.

c

Let's start by adding $3$ to both sides. In order to get rid of the denominator, we need to multiply both sides by $x.$

x-3=\dfrac{2}{x}-3

\text{LHS}+3=\text{RHS}+3

x=\dfrac{2}{x}

\text{LHS} \cdot x=\text{RHS}\cdot x

x^2=2

\sqrt{\text{LHS}}=\sqrt{\text{RHS}}

x=\pm \sqrt{2}

Both $x=\text{-} \sqrt{2}$ and $x=\sqrt{2}$ are solutions to the equation.

Solving Quadratic Equations with Square Roots 1.21 - Solution