# How do you solve -2x^2 - 7x + 4=0 using completing the square?

Jun 16, 2015

The answers are $x = \frac{1}{2}$ and $x = - 4$.

#### Explanation:

First, factor the coefficient of ${x}^{2}$ out of the first two terms to get $- 2 {x}^{2} - 7 x + 4 = - 2 \left({x}^{2} + \frac{7}{2} x\right) + 4 = 0$.

Next, take the coefficient of $x$ inside the parentheses, $\frac{7}{2}$, divide it by 2 to get $\frac{7}{4}$, and then square that number to get $\frac{49}{16}$. Add this number inside the parentheses and then "balance" it by adding $- 2 \cdot \frac{49}{16}$ on the other side of the equation to get $- 2 \left({x}^{2} + \frac{7}{2} x + \frac{49}{16}\right) + 4 = - 2 \cdot \frac{49}{16} = - \frac{49}{8}$.

The reason this trick is a good idea is that the expression ${x}^{2} + \frac{7}{2} x + \frac{49}{16}$ is a perfect square. It equals ${\left(x + \frac{7}{4}\right)}^{2}$, so the equation becomes $- 2 {\left(x + \frac{7}{4}\right)}^{2} + 4 = - \frac{49}{8}$, which is equivalent to $- 2 {\left(x + \frac{7}{4}\right)}^{2} = - \frac{81}{8}$ and ${\left(x + \frac{7}{4}\right)}^{2} = \frac{81}{16}$.

Now take the $\setminus \pm$ square root of both sides to get $x + \frac{7}{4} = \setminus \pm \frac{9}{4}$, leading to two solutions $x = \frac{9}{4} - \frac{7}{4} = \frac{2}{4} = \frac{1}{2}$ and $x = - \frac{9}{4} - \frac{7}{4} = - \frac{16}{4} = - 4$.

You should check these in the original equation:

$x = \frac{1}{2} \setminus R i g h t a r r o w - 2 {\left(\frac{1}{2}\right)}^{2} - 7 \left(\frac{1}{2}\right) + 4 = - \frac{1}{2} - \frac{7}{2} + 4 = - 4 + 4 = 0$

$x = - 4 \setminus R i g h t a r r o w - 2 {\left(- 4\right)}^{2} - 7 \left(- 4\right) + 4 = - 32 + 28 + 4 = 0$