# How do you solve the equation x^2-3/2x-23/16=0 by completing the square?

Nov 13, 2016

$x = \frac{3}{4} + \sqrt{2}$ or $x = \frac{3}{4} - \sqrt{2}$

#### Explanation:

${x}^{2} - \frac{3}{2} x - \frac{23}{16} = 0$

completing square of form ${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$, above becomes

${x}^{2} - 2 \times \frac{3}{4} \times x + {\left(\frac{3}{4}\right)}^{2} - {\left(\frac{3}{4}\right)}^{2} - \frac{23}{16} = 0$

or ${\left(x - \frac{3}{4}\right)}^{2} - \frac{9}{16} - \frac{23}{16} = 0$ or ${\left(x - \frac{3}{4}\right)}^{2} - \frac{32}{16} = 0$

or ${\left(x - \frac{3}{4}\right)}^{2} - {\left(\sqrt{2}\right)}^{2} = 0$

and factorizing it using ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, we get

$\left(x - \frac{3}{4} - \sqrt{2}\right) \left(x - \frac{3}{4} + \sqrt{2}\right) = 0$

Hence, either $x - \frac{3}{4} - \sqrt{2} = 0$ i.e. $x = \frac{3}{4} + \sqrt{2}$

or $x - \frac{3}{4} + \sqrt{2} = 0$ i.e. $x = \frac{3}{4} - \sqrt{2}$