# How do you solve x^2-2x-12=0 algebraically?

Apr 5, 2016

$x \approx 4.606 \text{ } - 2.606$ to 3 decimal places

$\textcolor{g r e e n}{x = 1 \pm \sqrt{13}}$

#### Explanation:

The whole number factors of 12 are {6,2}, {3,4} and none of these have a difference of 2 ( from $2 x$). So it must have none integer solutions for $x$. Thus we need to use the formulas. These formula are something worth remembering.

It takes quite a bit of work to make them stick!

Standard equation form:$\text{ } y = a {x}^{2} + b x + c$

where $\textcolor{m a \ge n t a}{x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}}$ ............................(1)

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Another one worth remembering is the completed square format.
Otherwise known as Vertex Form.

$\text{ } \textcolor{m a \ge n t a}{y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - \left[{\left(\frac{b}{2}\right)}^{2}\right]}$ .............................(2)

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$\textcolor{b l u e}{\text{Using equation (1)}}$

$\implies x \text{ "=" } \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(1\right) \left(- 12\right)}}{2 \left(1\right)}$

$\implies x \text{ "=" } \frac{2 \pm \sqrt{52}}{2}$

$\implies x \text{ "=" } \frac{2 \pm \sqrt{{2}^{2} \times 13}}{2}$

$\textcolor{g r e e n}{\implies x = 1 \pm \sqrt{13}}$
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$\textcolor{b l u e}{\text{Using equation (2)}}$

$y = 1 {\left(x - 1\right)}^{2} - 12 - 1$

$y = {\left(x - 1\right)}^{2} - 13$

Set $y = 0$

$0 = {\left(x - 1\right)}^{2} - 13$

$13 = {\left(x - 1\right)}^{2}$

Square root both sides

$\pm \sqrt{13} = x - 1$

$\textcolor{g r e e n}{x = 1 \pm \sqrt{13}}$
$x \approx 4.606 \text{ } - 2.606$ to 3 decimal places