# How do you solve 2x^2 + 1 = 4x?

Aug 7, 2015

$\left(1 + \frac{\sqrt{2}}{2}\right)$, $\left(1 - \frac{\sqrt{2}}{2}\right)$

#### Explanation:

$2 {x}^{2} + 1 = 4 x$

Therefore: $2 {x}^{2} - 4 x + 1 = 0$

If $a {x}^{2} + b x + c = 0$

Then ${x}_{1} = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}$

And ${x}_{2} = \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case $a = 2$, $b = - 4$, $c = 1$

${x}_{1} = \frac{- \left(- 4\right) + \sqrt{{\left(- 4\right)}^{2} - 4 \cdot \left(2\right) \cdot \left(1\right)}}{2 \cdot \left(2\right)}$

$= \frac{4 + \sqrt{16 - 8}}{4}$

$= \frac{4 + \sqrt{8}}{4}$

$= \frac{4 + \sqrt{2 \cdot 2 \cdot 2}}{4}$

$= \frac{4 + 2 \sqrt{2}}{4}$

$= 1 + \frac{\sqrt{2}}{2}$

Similarly:

${x}_{2} = 1 - \frac{\sqrt{2}}{2}$