# How do you find the domain of f(x)=sqrt(2x^2-5x+2)?

Apr 10, 2017

$D = R$ without $\left\{\frac{1}{2} , 2\right\}$

#### Explanation:

$2 {x}^{2} - 5 x + 2 \ge 0$

${b}^{2} - 4 a c = 25 - 4 \cdot 2 \cdot 2 = 9$
$\sqrt{9} = 3$

${x}_{1} = \frac{5 - 3}{4} = \frac{1}{2}$
${x}_{2} = \frac{5 + 3}{4} = 2$

$D = R$ without $\left\{\frac{1}{2} , 2\right\}$