# How do you use the discriminant to determine the nature of the solutions given  -3x^2-6x+15=0?

Jul 25, 2017

Roots are real , $x \approx - 3.45 , x \approx 1.45$

#### Explanation:

$- 3 {x}^{2} - 6 x + 15 = 0 \mathmr{and} - {x}^{2} - 2 x + 5 = 0$ . Comparing with standard

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

$a = - 1 , b = - 2 , c = 5$. Discriminant is

$D = {b}^{2} - 4 a c = {\left(- 2\right)}^{2} - 4 \cdot \left(- 1\right) \cdot 5 = 24$. If discriminant is

D> 0 ; 2 roots are real , if D= 0 ; 2 roots are equal ,

if D < 0 ; 2 roots are complex conjugate in nature. So here

$D > 0$ ,the roots are real.

Roots are $x = - \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$ or

$x = \frac{2}{-} 2 \pm \frac{\sqrt{24}}{-} 2 a = - 1 \pm \sqrt{6}$ or

$x \approx - 3.45 , x \approx 1.45$ [Ans]