How do you solve using the completing the square method 2x^2-4x-14=0?

2 Answers
Apr 28, 2018

The two solutions are x=+-2sqrt2+1.

Explanation:

First, divide the whole equation by two:

2x^2-4x-14=0

x^2-2x-7=0

Then, move the constant to the other side:

x^2-2x=7

Next, identify the square binomial on the left side. We know that (x-1)^2 expands to x^2-2x+1, so if we add 1 to both sides, we can use this backwards:

x^2-2xcolor(red)+color(red)1=7color(red)+color(red)1

(x-1)^2=8

Now, square root both sides:

x-1=+-sqrt8

x-1=+-2sqrt2

x=+-2sqrt2+1

Those are the solutions. Hope this helped!

Apr 28, 2018

x=1pm2sqrt2

Explanation:

Factor out 2:

2[x^2-2x]-14 lArr notice we didn't factor the constant.

Complete the square of the brackets:

2[(x-1)^2-1]-14

Multiply the brackets:

2(x-1)^2-2-14

Simplify:

2(x-1)^2-16

We can always check this:

(x-1)^2=x^2-2x+1

2(x^2-2x+1)=2x^2-4x+2-16 -> 2x^2-4x-14

Solving:

2(x-1)^2-16=0

2(x-1)^2=16

(x-1)^2=8

x-1=pmsqrt8

x=1pmsqrt8

-> x=1pm2sqrt2