# Given 2x^2-1=2x, how do you find the discriminant and the number of solutions?

Mar 29, 2018

Solution: $x = \frac{1}{2} + \frac{\sqrt{3}}{2} \mathmr{and} x = \frac{1}{2} - \frac{\sqrt{3}}{2}$

#### Explanation:

$2 {x}^{2} - 1 = 2 x \mathmr{and} 2 {x}^{2} - 2 x - 1 = 0$ Comparing with standard

quadratic equation $a {x}^{2} + b x + c = 0$ we get,

$a = 2 , b = - 2 , c = - 1$ Discriminant $D = {b}^{2} - 4 a c$ or

$D = 4 + 8 = 12$ If discriminant positive, we get two real

solutions, if it is zero we get just one solution, and if it is negative

we get complex solutions.Discriminant is positive here , so it has

two real roots . Quadratic formula: $x = \frac{- b \pm \sqrt{D}}{2 a}$or

$x = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2 \sqrt{3}}{4} = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$

Solution: $x = \frac{1}{2} + \frac{\sqrt{3}}{2} \mathmr{and} x = \frac{1}{2} - \frac{\sqrt{3}}{2}$ [Ans]