# How do you find the value of the discriminant and state the type of solutions given 5b^2+b-2=0?

Sep 2, 2017

$D = 41$ , two real solutions and $b \approx - 0.74 , b \approx 0.54$.

#### Explanation:

$5 {b}^{2} + b - 2 = 0$ . Comparing with standard quadratic equation

$a {x}^{2} + b x + c = 0$ we get here $a = 5 , b = 1 , c = - 2$

Discriminant $D = {b}^{2} - 4 \cdot a \cdot c = 1 - 4 \cdot 5 \cdot \left(- 2\right) = 1 + 40 = 41$

If $D$ is positive, we will get two real solutions, if it is zero we get

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

Here $D$ is positive , so there are two real solution.

$b = \frac{- b \pm \sqrt{D}}{2 a} \mathmr{and} b = \frac{- 1 \pm \sqrt{41}}{10}$ or

$\therefore b \approx - 0.74 \left(2 \mathrm{dp}\right) , b \approx 0.54 \left(2 \mathrm{dp}\right)$ [Ans]