How do you solve using the completing the square method 2x2−4x−14=0?

2 Answers
Aviv S.
Apr 28, 2018

The two solutions are x=±2√2+1.

Explanation:

First, divide the whole equation by two:

2x2−4x−14=0

x2−2x−7=0

Then, move the constant to the other side:

x2−2x=7

Next, identify the square binomial on the left side. We know that (x−1)2 expands to x2−2x+1, so if we add 1 to both sides, we can use this backwards:

x2−2x+1=7+1

(x−1)2=8

Now, square root both sides:

x−1=±√8

x−1=±2√2

x=±2√2+1

Those are the solutions. Hope this helped!

Brandon
Apr 28, 2018

x=1±2√2

Explanation:

Factor out 2:

2[x2−2x]−14 ⇐ 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=x2−2x+1

2(x2−2x+1)=2x2−4x+2−16→2x2−4x−14

Solving:

2(x−1)2−16=0

2(x−1)2=16

(x−1)2=8

x−1=±√8

x=1±√8

→x=1±2√2

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