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Completing the Square

Completing the Square 1.13 - Solution

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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 xx are on one side of the equation and the constant is on the other side. 2x2+6x12=02x2+6x=12\begin{gathered} 2x^2+6x-12=0\quad\Leftrightarrow\quad 2x^2+6x=12 \end{gathered} Now let's divide each side by 22 so the coefficient of x2x^2 will be 1.1.
2x2+6x=122x^2+6x=12
x2+3x=6x^2+3x=6
In a quadratic expression, bb is the coefficient of the xx-term. For the equation above, we have that b=3.b=3. Let's now calculate (b2)2.\left( \frac{b}{2} \right)^2.
(b2)2\left( \dfrac{b}{2} \right)^2
(32)2\left( \dfrac{{\color{#0000FF}{3}}}{2} \right)^2
94\dfrac{9}{4}
Next, we will add (b2)2=94\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.
x2+3x=6x^2+3x=6
x2+3x+94=6+94x^2+3x+{\color{#0000FF}{\dfrac{9}{4}}}=6+{\color{#0000FF}{\dfrac{9}{4}}}
(x+32)2=6+94\left(x+\dfrac{3}{2}\right)^2=6+\dfrac{9}{4}
Simplify right-hand side
(x+32)2=244+94\left(x+\dfrac{3}{2}\right)^2=\dfrac{24}{4}+\dfrac{9}{4}
(x+32)2=334\left(x+\dfrac{3}{2}\right)^2=\dfrac{33}{4}
Simplify left-hand side
(x+32)2=334\sqrt{\left(x+\dfrac{3}{2}\right)^2}=\sqrt{\dfrac{33}{4}}
(x+32)2=334\sqrt{\left(x+\dfrac{3}{2}\right)^2}=\dfrac{\sqrt{33}}{\sqrt{4}}
x+32=±332x+\dfrac{3}{2}=\pm \dfrac{\sqrt{33}}{2}
x=-32±332x=\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=-32±332x=\text{-} \dfrac{3}{2} \pm \dfrac{\sqrt{33}}{2}
x=-32332x=\text{-} \dfrac{3}{2} - \dfrac{\sqrt{33}}{2} x=-32+332x=\text{-} \dfrac{3}{2} + \dfrac{\sqrt{33}}{2}
x-4.37x \approx \text{-} 4.37 x1.37x \approx 1.37