# How do you find the discriminant of 2x^2-3x+1=0 and use it to determine if the equation has one, two real or two imaginary roots?

Aug 1, 2017

Two real roots $\frac{1}{2}$ and $1$.

#### Explanation:

The discriminant of the quadratic equation $a {x}^{2} + b x + c = 0$ is ${b}^{2} - 4 a c$.

Assuming $a$, $b$ and $c$ are real numbers,

if ${b}^{2} - 4 a c > 0$, we have two real roots

if ${b}^{2} - 4 a c = 0$, we have one real root, and

if ${b}^{2} - 4 a c < 0$, we have two complex conjugate numbers as roots.

For $2 {x}^{2} - 3 x + 1 = 0$, as $a = 2$, $b = - 3$ and $c = 1$,

the discriminant is ${\left(- 3\right)}^{2} - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 > 0$,

hence we should have two real roots

Now $2 {x}^{2} - 3 x + 1 = 0$ can be written as

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

or $2 x \left(x - 1\right) - 1 \left(x - 1\right) = 0$ or $\left(2 x - 1\right) \left(x - 1\right) = 0$

i.e. $x = \frac{1}{2}$ or $1$, i.e. two real roots.