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

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Answer 1 · 2017-07-25T11:31:42

Roots are real , #x ~~-3.45 , x ~~1.45#

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

quadratic equation # ax^2+bx+c=0 # we get here

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

#D = b^2-4ac =( -2)^2-4*(-1)*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 = -b/(2a) +- sqrt (b^2-4ac)/(2a) # or

# x = 2/-2 +- sqrt (24)/-2a = -1 +- sqrt6 # or

#x ~~-3.45 , x ~~1.45# [Ans]