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

1 Answer
Mar 3, 2016

Answer:

Multiply out and simplify to find a quadratic in standard form, then apply the formula to find zeros #x = 3 +-i#

Explanation:

#y=(x-2)^2+(x-4)^2#

#=(x^2-4x+4)+(x^2-8x+16)#

#=2x^2-12x+20#

#=2(x^2-6x+10)#

#x^2-6x+10# is in the form #ax^2+bx+c# with #a=1#, #b=-6# and #c = 10#.

This has zeros given by the quadratic formula:

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

#=(6+-sqrt((-6)^2-(4*1*10)))/(2*1)#

#=(6+-sqrt(36-40))/2#

#=(6+-sqrt(-4))/2#

#=(6+-sqrt(4)i)/2#

#=(6+-2i)/2#

#=3+-i#