How do you find the roots, real and imaginary, of #y= 2(x+5)^2+(x-4)^2 # using the quadratic formula?

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Answer 1 · 2016-09-03T11:32:51

Zeros are #-2-3sqrt2i# and #-2+3sqrt2i#

Quadratic formula gives the roots of #y=ax^2+bx+c# as #x=(-b+-sqrt(b^2-4ac))/(2a)#

As #y=2(x+5)^2+(x-4)^2#

= #2(x^2+10x+25)+x^2-8x+16#

= #2x^2+20x+50+x^2-8x+16#

= #3x^2+12x+66#

Hence zeros are given by

#x=(-12+-sqrt(12^2-4×3×66))/(2×3)#

= #(-12+-sqrt(144-792))/6#

= #(-12+-sqrt(-648))/6#

= #(-12+-18sqrt2i)/6#

= #-2+-3sqrt2i#

Hence, zeros are #-2-3sqrt2i# and #-2+3sqrt2i#