Question

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

Answer 1

See below.

Explanation:

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

`ax^2+bx+c=0`

And turn it into

`a(x+d)^2+e=0`

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

`y=-2(x^2-2x-7/2)`

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

`y=-2(x^2color(blue)(-2)x-7/2)`

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

`(-2/2)^2=(1)^2=1`

I will rewrite the equation:

`y=-2(x^2-2x+f-7/2-f)`

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

`=>y=-2(x^2-2x+1-7/2-1)`

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

`=>y=-2[(x^2-2x+1)-7/2-1]`

Simplify:

`=>y=-2[(x^2-2x+1)-9/2]`

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

`=>y=-2[(x-1)^2-9/2]`

Distribute `-2`:

`=>y=-2(x-1)^2+9`

Or, equivalently:

`y=9-2(x-1)^2`