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

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Answer 1 · 2017-07-25T11:31:42

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

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

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

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

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

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

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

$D > 0$ $D>0$ ,the roots are real.

Roots are $x = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}$ $x = -b/(2a) +- sqrt (b^2-4ac)/(2a)$ or

$x = \frac{2}{- 2} \pm \frac{\sqrt{24}}{- 2} a = - 1 \pm \sqrt{6}$ $x = 2/-2 +- sqrt (24)/-2a = -1 +- sqrt6$ or

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

How do you use the discriminant to determine the nature of the solutions given -3x^2-6x+15=0−3x2−6x+15=0 -3x^2-6x+15=0?