How do you find the roots, real and imaginary, of #y=-x^2+5x-(x-3)^2 # using the quadratic formula?

1 Answer
Apr 25, 2018

Answer:

#x=1or4.5#

Explanation:

First, we expand #-(x-3)^2# to get #-(x^2-6x+9)=-x^2+6x-9#

#y=-x^2+5x-x^2+6x-9=-2x^2+11x-9#

#x=(-11+-sqrt(11^2-4(-2*-9)))/(2*-2)#

#x=-(-11+-sqrt(121-4(18)))/4#

#x=-(-11+-sqrt(49))/4#

#x=-(-11+7)/4or-(-11-7)/4#

#x=4/4or18/4#

#x=1or4.5#

Graph:
graph{-x^2+5x-(x-3)^2 [-2.045, 7.955, -0.68, 4.32]}