Question

How do you find the roots, real and imaginary, of `y= 12x^2 - 12x + 8+ 4(-x-1)^2` using the quadratic formula?

Answer 1

Two imaginary roots.
`x = (1 +- isqrt47)/8`

Explanation:

Develop and bring the equation to standard form:
`4(-x - 1)^2 = 4[-(x + 1)]^2 = 4(x + 1)^2 = 4(x^2 + 2x + 1) = 4x^2 + 8x + 4`.
`y = 12x^2 - 12x + 8 + (4x^2 + 8x + 4) = 0`
`y = 16x^2 - 4x + 12 = 4(4x^2 - x + 3) = 0`
Solve the quadratic equation by the improved quadratic formula (Socratic Search).
`y = 4x^2 - x + 3 = 0`
`D = 1 - 48 = -47 < 0` --> `d = +- isqrt47`
Since D < 0, there are 2 imaginary roots.
`x = -b/(2a) +- d/(2a) = 1/8 +- (isqrt47)/8 = (1 +- isqrt47)/8`

Answer 2

`x=1/8+-sqrt(47)/8color(white)(.)i`

Explanation:

Given:`" "y=12x^2-12x+8+4(-x-1)^2`

Squaring the bracket

`y=12x^2-12x+8+4(x^2+2x+1)`

giving:

`y=12x^2-12x+8+4x^2+8x+4`

`y=16x^2-4x+12 color(red)(larr" slightly different approach hear.")`

Set `y=0`

`=>0=16x^2-4x+12`

divide both sides by 4

Note that `0/4=0`

`=>0=4x^2-x+3`

Compare to `y=ax^2+bx+c " "->" "x=(-b+-sqrt(b^2-4ac))/(2a)`

Where`" "a=4"; "b=-1"; "c=3`

`=>x=(1+-sqrt((-1)^2-4(4)(3)))/(2(4))`

`x=(1+-sqrt(1-48))/8`

`x=1/8+-sqrt(-47)/8`

But 47 is a prime number so we have

`x=1/8+-sqrt(47)/8color(white)(.)i`