# How do you find the discriminant and how many solutions does 2x^2-4x+1=0 have?

Apr 30, 2018

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

has two unequal real roots

#### Explanation:

the discriminant
for

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

is

$\Delta = {b}^{2} - 4 a c$

$\Delta > 0 \implies \text{ two unequal real roots}$

$\Delta = 0 \implies \text{ equal real roots (ie one root)}$

$\Delta < 0 \implies \text{ two complex roots}$

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

$a = 2 , b = - 4 , c = 1$

$\Delta = {\left(- 4\right)}^{2} - 4 \times 2 \times 1$

$\Delta = 16 - 8 = 8$

$\therefore \Delta > 0$

we conclude that

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

has two unequal real roots