# How do you solve 16x^2 - 8x + 1 = 49?

Oct 29, 2015

Solve $16 {x}^{2} - 8 x + 1 = 47$

Ans: 2 and $- \frac{3}{2}$

#### Explanation:

The two sides are both perfect squares.
$16 {x}^{2} - 8 x + 1 = {\left(4 x - 1\right)}^{2} = {7}^{2}$
$\left(4 x - 1\right) = \pm 7$

4x - 1 = 7 --> 4x = 8 --> $x = 2$

4x - 1 = -7 --> 4x = -6 --> $x = - \frac{6}{4} = - \frac{3}{2}$