How do you find the roots, real and imaginary, of #y=x^2 + 14x +2(x/2-2)^2 # using the quadratic formula?

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Answer 1 · 2016-02-17T15:53:41

#-0.382# and #-2.618#

First distribute the 2 to the binomial.

#y=x^2+14x+2(x/2-2)^2#

#y=x^2+14x+(x-4)^2#

Next, square #(x-4)^2# using FOIL.

#x^2+14x+x^2-8x+16#

Combine like terms.

#2x^2+6x+16#

Divide all terms by #2#

#x^2+3x+8#

Now, using the quadratic formula #(-b+-sqrt(b^2-4))/2a# where #a=1, b=3 and c=8#

This gives #(-3+-sqrt(3^2-4))/(2*1)#

==> #(-3+-sqrt(9-4))/2#

==> #(-3+-sqrt5)/2#

This may be a good answer. If not, I'll use decimals...

==> #(-3+-2.236)/2#

==> #(-3+2.236)/2# = #-0.382#

==> #(-3-2.236)/2# = #-2.618#