Question

How do you find the roots, real and imaginary, of `y= (3x+4)^2-13x-x^2+14` using the quadratic formula?

Answer 1

THere are two complex conjugate roots:

`x = -11/16 +- sqrt(839)/16i`

Explanation:

We have:

`y = (3x+4)^2-13x-x^2+14`

We seek the roots of `y=0`, ie of:

`(3x+4)^2-13x-x^2+14 = 0`

Expanding we get:

`9x^2+24x+16-13x-x^2+14 = 0`
`:. 8x^2+11x+30 = 0`

Using the quadratic formula with:

`a=8, b=11, c=30`

we have:

`x = (-b +- sqrt(b^2-4ac))/(2a)`
`\ \ = (-11 +- sqrt(11^2-4.8.30))/(2.8)`
`\ \ = (-11 +- sqrt(121-960))/(16)`
`\ \ = (-11 +- sqrt(-839))/(16)`
`\ \ = -11/16 +- sqrt(839)/16i`