How do you find the roots, real and imaginary, of #y= -3x^2 -5x +3(x/2+1)^2 # using the quadratic formula?

1 Answer
Feb 23, 2018

Answer:

#(-4+-2sqrt(31))/(9)#

Explanation:

To solve for roots, we have to substitute y as 0.

#-3x^2-5x+3(x/2+1)^2=0#
#-3x^2-5x+3(x^2/4+x+1)=0#
#-3x^2-5x+(3x^2)/4+3x+3=0#
#-12x^2-20x+3x^2+12x+12=0#
#-9x^2-8x+12=0#

#9x^2+8x-12=0#

First, let confirm whether the roots are real or imaginary.
#delta=8^2-4(9)(-12)=496>0#
Hence, both roots are real.

For a quadratic equation #Ax^2+Bx+C=0#, the roots are
#(-B+-sqrt(B^2-4AC))/(2A)#

Therefore the req. roots
#=(-8+-sqrt(delta))/(2*9)#

#=(-8+-4sqrt(31))/(18)#

#=(-4+-2sqrt(31))/(9)#

Note:
1. #delta# refers to the value in the square root in the quadratic equation.
2. #+-# means that #+# and #-# can contribute to two equally correct answers, in this case, the two roots are actually #(-4+2sqrt(31))/(9)# and #(-4-2sqrt(31))/(9)#