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
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
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#
