How do you find the roots, real and imaginary, of y=-(2x+2)^2+x^2 + 9x-1 using the quadratic formula?

1 Answer
Sep 6, 2017

The roots are imaginary. y=[1+isqrt(59)]/6 or y=[1-isqrt(59)]/6

Explanation:

Firstly, expand

y= -[4x^2+8x+4]+x^2+9x-1

= -4x^2-8x-4+x^2+9x-1

Simplifying

y= -3x^2+x-5

Using [-b+-sqrt(b^2-4ac)]/(2a)

Here

a=-3, b=1, c=-5

Then,

y=[-1+-sqrt(1^2-4*3*5)]/(2*-3)

= [cancel(-)1+-sqrt(1-60)]/cancel(-)6

= [1+-sqrt(-59)]/6

= [1+-sqrt(59)i]/6

So the roots are:

y=[1+isqrt(59)]/6 or y=[1-isqrt(59)]/6

So the roots are imaginary.