How do you find the roots, real and imaginary, of $y=2x^2 + 4x +4(x/2-1)^2$ using the quadratic formula?

Question

1505 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-09T10:33:42

$x=(2isqrt3)/3, -(2isqrt3)/3$

We can first expand the bracket:

$y=2x^2+4x+4(x/2-1)^2$

$y=2x^2+4x+4(x^2/4-x/2-x/2+1)$

and now distribute the 4:

$y=2x^2+4x+4(x^2/4-x+1)$

$y=2x^2+4x+x^2-4x+4$

combine terms:

$y=3x^2+4$

and now see that we can use the quadratic formula, with values $a=3, b=0, c=4$

$x = (-b \pm sqrt(b^2-4ac)) / (2a)$

$x = (0 \pm sqrt(0^2-4(3)(4))) / (2(3))$

$x = ( \pm sqrt(-48)) / 6$

$x = ( \pm4i sqrt(3)) / 6 =pm(2isqrt3)/3$

How do you find the roots, real and imaginary, of y=2x^2 + 4x +4(x/2-1)^2 y=2x^2 + 4x +4(x/2-1)^2 using the quadratic formula?