How do you solve y=-2x^2+4x+7 using the completing square method?

Dec 31, 2016

See below.

Explanation:

To complete the square, we take a quadratic equation of the form

$a {x}^{2} + b x + c = 0$

And turn it into

$a {\left(x + d\right)}^{2} + e = 0$

Begin by factoring out $- 2$ to get a coefficient of $1$ for the ${x}^{2}$ term.

$y = - 2 \left({x}^{2} - 2 x - \frac{7}{2}\right)$

Now look at the coefficient of the $x$ term.

$y = - 2 \left({x}^{2} \textcolor{b l u e}{- 2} x - \frac{7}{2}\right)$

Divide this coefficient by $2$ and square the result:

${\left(- \frac{2}{2}\right)}^{2} = {\left(1\right)}^{2} = 1$

I will rewrite the equation:

$y = - 2 \left({x}^{2} - 2 x + f - \frac{7}{2} - f\right)$

Replace $f$ with the result of the above operation:

$\implies y = - 2 \left({x}^{2} - 2 x + 1 - \frac{7}{2} - 1\right)$

We separate off the first part of the parentheses from the second:

$\implies y = - 2 \left[\left({x}^{2} - 2 x + 1\right) - \frac{7}{2} - 1\right]$

Simplify:

$\implies y = - 2 \left[\left({x}^{2} - 2 x + 1\right) - \frac{9}{2}\right]$

What we have left in the parentheses is a perfect square. Factor:

$\implies y = - 2 \left[{\left(x - 1\right)}^{2} - \frac{9}{2}\right]$

Distribute $- 2$:

$\implies y = - 2 {\left(x - 1\right)}^{2} + 9$

Or, equivalently:

$y = 9 - 2 {\left(x - 1\right)}^{2}$