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

1 Answer
Apr 24, 2016

Answer:

#x = 7/2+-1/2i#

Explanation:

First multiply out the quadratic to get it into standard form:

#y = (x-3)^2+(x-4)^2#

#=x^2-6x+9+x^2-8x+16#

#=2x^2-14x+25#

This is now in the form #ax^2+bx+c# with #a=2#, #b=-14# and #c=25#

This has zeros given by the quadratic formula:

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

#=(14+-sqrt((-14)^2-4(2)(25)))/(2*2)#

#=(14+-sqrt(196-200))/4#

#=(14+-sqrt(-4))/4#

#=(14+-sqrt(4)i)/4#

#=7/2+-1/2i#