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...

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

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

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

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