# How do you solve 9x^2-12x-14=0 by completing the square?

Aug 13, 2016

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

#### Explanation:

In $9 {x}^{2} - 12 x - 14 = 0$, while $9 {x}^{2} = {\left(3 x\right)}^{2}$, to complete the square, recall the identity ${\left(x \pm a\right)}^{2} = {x}^{2} \pm 2 a x + {a}^{2}$.

As -12x=-2×(3x)×2, we need to add ${2}^{2}$ to make it complete square.

Hence, $9 {x}^{2} - 12 x - 14 = 0$ can be written as ((3x)^2-2×(3x)×2+2^2)-4-14=0 or

${\left(3 x - 2\right)}^{2} - 18 = 0$, which is equivalent to

${\left(3 x - 2\right)}^{2} - {\left(\sqrt{18}\right)}^{2} = 0$

and using identity ${\left(a - b\right)}^{2} = \left(a + b\right) \left(a - b\right)$ we can write the equation as

$\left(3 x - 2 + \sqrt{18}\right) \left(3 x - 2 - \sqrt{18}\right) = 0$

i.e. either $3 x - 2 + \sqrt{18} = 0$ or $3 x - 2 - \sqrt{18} = 0$.

Now as $\sqrt{18} = 3 \sqrt{2}$

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