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Completing the Square

Completing the Square 1.13 - Solution

We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so all terms with $x$ are on one side of the equation and the constant is on the other side. $\begin{gathered} 2x^2+6x-12=0\quad\Leftrightarrow\quad 2x^2+6x=12 \end{gathered}$ Now let's divide each side by $2$ so the coefficient of $x^2$ will be $1.$
$2x^2+6x=12$
$x^2+3x=6$
In a quadratic expression, $b$ is the coefficient of the $x$-term. For the equation above, we have that $b=3.$ Let's now calculate $\left( \frac{b}{2} \right)^2.$
$\left( \dfrac{b}{2} \right)^2$
$\left( \dfrac{{\color{#0000FF}{3}}}{2} \right)^2$
$\dfrac{9}{4}$
Next, we will add $\left( \frac{b}{2} \right)^2=\frac{9}{4}$ to both sides of our equation. Then, we will factor the trinomial on the left-hand side, and solve the equation.
$x^2+3x=6$
$x^2+3x+{\color{#0000FF}{\dfrac{9}{4}}}=6+{\color{#0000FF}{\dfrac{9}{4}}}$
$\left(x+\dfrac{3}{2}\right)^2=6+\dfrac{9}{4}$
Simplify right-hand side
$\left(x+\dfrac{3}{2}\right)^2=\dfrac{24}{4}+\dfrac{9}{4}$
$\left(x+\dfrac{3}{2}\right)^2=\dfrac{33}{4}$
Simplify left-hand side
$\sqrt{\left(x+\dfrac{3}{2}\right)^2}=\sqrt{\dfrac{33}{4}}$
$\sqrt{\left(x+\dfrac{3}{2}\right)^2}=\dfrac{\sqrt{33}}{\sqrt{4}}$
$x+\dfrac{3}{2}=\pm \dfrac{\sqrt{33}}{2}$
$x=\text{-} \dfrac{3}{2} \pm \dfrac{\sqrt{33}}{2}$
We will now find the first and second solutions by using the positive and negative signs.
$x=\text{-} \dfrac{3}{2} \pm \dfrac{\sqrt{33}}{2}$
$x=\text{-} \dfrac{3}{2} - \dfrac{\sqrt{33}}{2}$ $x=\text{-} \dfrac{3}{2} + \dfrac{\sqrt{33}}{2}$
$x \approx \text{-} 4.37$ $x \approx 1.37$