# How do you solve by completing the square x^2+3x-8=0?

Apr 3, 2015

Solving a quadratic expression by completing the square means to manipulate the expression in order to write it in the form
${\left(x + a\right)}^{2} = b$
So, if $b \setminus \ge 0$, you can take the square root at both sides to get
$x + a = \setminus \pm \setminus \sqrt{b}$
and conclude $x = \setminus \pm \setminus \sqrt{b} - a$.

Now, we have ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$. Since you equation starts with ${x}^{2} + 3 x$, this means that $2 a x = 3 x$, and so $a = \frac{3}{2}$.
Adding $\frac{41}{4}$ at both sides, we have
${x}^{2} + 3 x + \frac{9}{4} = \frac{41}{4}$
Which is the form we wanted, because now we have
${\left(x + \frac{3}{2}\right)}^{2} = \frac{41}{4}$
$x + \frac{3}{2} = \setminus \pm \setminus \sqrt{\frac{41}{4}} = \setminus \pm \setminus \frac{\sqrt{41}}{2}$ and finally $x = \setminus \pm \setminus \frac{\sqrt{41}}{2} - \frac{3}{2}$