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

Jul 15, 2016

$x = - 4$ or $x = \frac{1}{2}$

#### Explanation:

As we have to solve the quadratic equation by completing square method and $2 {x}^{2}$ is not a square, let us first divide $2 {x}^{2} + 7 x - 4 = 0$ by $2$, which gives us

${x}^{2} + \frac{7}{2} x - 2 = 0$. ....(A)

Now recall the identity ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$ which shows that to complete square we must add square of half of $x ' s$ coefficient. As coefficient of $x$ is $\frac{7}{2}$, we add and subtract ${\left(\frac{7}{4}\right)}^{2} = \frac{49}{16}$ and equation (A) becomes

${x}^{2} + \frac{7}{2} x + \frac{49}{16} - \frac{49}{16} - 2 = 0$ or

${\left(x + \frac{7}{4}\right)}^{2} - \frac{49 + 32}{16} = 0$ or

${\left(x + \frac{7}{4}\right)}^{2} - \frac{81}{16} = 0$ or

${\left(x + \frac{7}{4}\right)}^{2} - {\left(\frac{9}{4}\right)}^{2} = 0$ and using identity $\left({x}^{2} - {a}^{2}\right) = \left(x + a\right) \left(x - a\right)$, we get

$\left(x + \frac{7}{4} + \frac{9}{4}\right) \left(x + \frac{7}{4} - \frac{9}{4}\right) = 0$ or

$\left(x + \frac{16}{4}\right) \left(x - \frac{2}{4}\right) = 0$

$\left(x + 4\right) \left(x - \frac{1}{2}\right) = 0$

i.e. $x = - 4$ or $x = \frac{1}{2}$