# How do i complete the square of this equation?

## $2 {x}^{2} + 3 x - 4 = 0$

Mar 23, 2017

Completing the square only works if the coefficient of the square term is 1, therefore, we divide both sides of the given equation by 2:

${x}^{2} + \frac{3}{2} x - 2 = 0$

Using the pattern ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$, we see that we must add ${a}^{2} + 2$ to both sides:

${x}^{2} + \frac{3}{2} x + {a}^{2} = {a}^{2} + 2$

Please notice that the left side of the equation, now, fits the right side of the pattern. We can use the middle term in the pattern and the middle term on the left side of the equation to find the value of "a" by setting the two terms equal:

$2 a x = \frac{3}{2} x$

Divide both sides by 2x:

$a = \frac{3}{4}$

We know that the left side of the equation becomes left side of the pattern with value of "a" and we substitute the value of "a" on the right side, too:

${\left(x + \frac{3}{4}\right)}^{2} = {\left(\frac{3}{4}\right)}^{2} + 2$

The square is completed.

Though you did not ask for it, I will continue and solve for x.

Simplify the right:

${\left(x + \frac{3}{4}\right)}^{2} = \frac{9}{16} + \frac{32}{16}$

${\left(x + \frac{3}{4}\right)}^{2} = \frac{41}{16}$

Square root both sides:

$x + \frac{3}{4} = \pm \frac{\sqrt{41}}{4}$

$x = - \frac{3}{4} \pm \frac{\sqrt{41}}{4}$