# How do you solve 2x^2-8x+5>=0?

Jul 31, 2016

Solution is $x \ge 2 + \sqrt{\frac{3}{2}}$ or $x \le 2 - \sqrt{\frac{3}{2}}$

#### Explanation:

$2 {x}^{2} - 8 x + 5 \ge 0$ is equivalent to

$2 \left({x}^{2} - 4 x + \frac{5}{2}\right) \ge 0$ or

$2 \left({x}^{2} - 4 x + 4 - 4 + \frac{5}{2}\right) \ge 0$ or

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

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

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

i.e. either $\left(x - 2 + \sqrt{\frac{3}{2}}\right) \ge 0$ and $\left(x - 2 - \sqrt{\frac{3}{2}}\right) \ge 0$ i.e.

$x \ge 2 - \sqrt{\frac{3}{2}}$ and $x \ge 2 + \sqrt{\frac{3}{2}}$ i.e. $x \ge 2 + \sqrt{\frac{3}{2}}$

or $\left(x - 2 + \sqrt{\frac{3}{2}}\right) \le 0$ and $\left(x - 2 - \sqrt{\frac{3}{2}}\right) \le 0$ i.e.

$x \le 2 - \sqrt{\frac{3}{2}}$ and $x \le 2 + \sqrt{\frac{3}{2}}$ i.e. $x \le 2 - \sqrt{\frac{3}{2}}$

Hence solution is $x \ge 2 + \sqrt{\frac{3}{2}}$ or $x \le 2 - \sqrt{\frac{3}{2}}$

graph{2x^2-8x+5 [-3, 7, -3.16, 1.84]}