# How do you use the discriminant to find the number of real solutions of the following quadratic equation: 2x^2 + 4x + 2 = 0?

Jun 27, 2015

${b}^{2} - 4 a c = {4}^{2} - 4 \left(2\right) \left(2\right) = 0$ so this equation has exactly one real solution.

#### Explanation:

For a quadratic of the form $a {x}^{2} + b x + c = 0$
the solution roots are given by the quadratic formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The discriminant is the portion $\left({b}^{2} - 4 a c\right)$
if it is negative then the roots include a term which is the square root of a negative number; since there are no Real numbers whose square root is negative, when the discriminant is $< 0$ there are no Real roots.

If the discriminant is $= 0$ then $\pm \sqrt{0}$ is a single value ($+ 0 = - 0$)
and the quadratic has exactly one Real solution.

If the discriminant is $> 0$ then $\pm \sqrt{\text{discriminant}}$ represents two different values and the quadratic has two Real solutions.