# How do you solve using the completing the square method 2x^2 - 7x - 15 = 0?

Mar 30, 2018

#### Answer:

$x = - \frac{3}{2} \text{ or } x = 5$

#### Explanation:

$\text{to use the method of "color(blue)"completing the square}$

• " the coefficient of the "x^2" term must be 1"

$\Rightarrow 2 \left({x}^{2} - \frac{7}{2} x - \frac{15}{2}\right) = 0$

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - \frac{7}{2} x$

$2 \left({x}^{2} + 2 \left(- \frac{7}{4}\right) x \textcolor{red}{+ \frac{49}{16}} \textcolor{red}{- \frac{49}{16}} - \frac{15}{2}\right) = 0$

$\Rightarrow 2 {\left(x - \frac{7}{4}\right)}^{2} + 2 \left(- \frac{49}{16} - \frac{15}{2}\right) = 0$

$\Rightarrow 2 {\left(x - \frac{7}{4}\right)}^{2} - \frac{169}{8} = 0$

$\Rightarrow 2 {\left(x - \frac{7}{4}\right)}^{2} = \frac{169}{8}$

$\Rightarrow {\left(x - \frac{7}{4}\right)}^{2} = \frac{169}{16}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\Rightarrow x - \frac{7}{4} = \pm \sqrt{\frac{169}{16}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = \frac{7}{4} \pm \frac{13}{4}$

$\Rightarrow x = \frac{7}{4} - \frac{13}{4} = - \frac{3}{2} \text{ or } x = \frac{7}{4} + \frac{13}{4} = 5$