How do you find the roots, real and imaginary, of #y=-(x-2)^2+x^2 + 11x +28 # using the quadratic formula?
1 Answer
Jan 28, 2017
Explanation:
Let's expand and simplify the given equation:
#y = -(x-2)^2+x^2+11x+28#
#color(white)(y) = -(x^2-4x+4)+x^2+11x+28#
#color(white)(y) = color(red)(cancel(color(black)(-x^2)))+4x-4+color(red)(cancel(color(black)(x^2)))+11x+28#
#color(white)(y) = 15x+24#
Since the resulting equation is linear, not quadratic, the quadratic formula does not apply.
This linear equation has one zero, namely:
#x = -24/15 = -(color(red)(cancel(color(black)(3)))*8)/(color(red)(cancel(color(black)(3)))*5) = -8/5#
