Question

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

Answer 1

`(-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)`