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

1 Answer
Dec 31, 2016

Answer:

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#