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

1 Answer
Jul 14, 2016

Answer:

# 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#