How do you find the roots, real and imaginary, of #y=-5(x-3)^2-(x-3)^2-10 # using the quadratic formula?

1 Answer
Apr 26, 2016

Answer:

#3-i10sqrt2# and #3+i10sqrt2# are roots of given equation.

Explanation:

For equation #ax^2+bx+c=0#, quadratic formula gives #x=(-b+-sqrt(b^2-4ac))/(2a)#.

In equation #y=-5(x-3)^2-(x-3)^2-10=-6(x-3)^2-10#. Hence, #(x-3)=(0+-sqrt(0^2-4(-5)(-10)))/(2xx(-5))=(+-sqrt(-200))/(-10)=+-i10sqrt2/(-10)=+-i10sqrt2#

Hence, #x=3+-i10sqrt2# and these are roots of given equation.