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

(-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)