Question

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

Answer 1

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!

Answer 2

`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`