Question

How do you find the roots, real and imaginary, of `y=(x+4)(x+1)-3x^2+2` using the quadratic formula?

Answer 1

`x = 5/4 +- (sqrt(73))/4`

Explanation:

Multiply out the bracket:

`(x+4)(x+1) = x^2 + 5x + 4`

`therefore y = -2x^2 + 5x + 6`

For quadratic `y = ax^2 + bx + c` the roots occur at `y=0` and are found with the quadratic formula. As it is a polynomial of order 2, we expect to get two roots.

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`x = (-5+-sqrt(25-4(-2)(6)))/(2(-2)) = (-5+-sqrt(73))/(-4)`

`therefore x = 5/4 +- (sqrt(73))/4`