I was covering "completing the square" in class today and one of the examples caused a little difficulty. I thought I would use it as a chance to test a new plug-in for this site.

The question involved was: Use completing the square to solve 2x2=3x+6.

We begin by rearranging and setting equal to zero

2x2−3x−6=0

and then remove the common factor of 2. I tend to remove it just from the first 2 terms to avoid fractions in the third term - this doesn't actually matter in this case but I will continue with the method I used in class

2x2−3x−6=2(x2−32)−6.

Now we can completing the square:

2(x2−32)−6=2[(x−34)2−916]−6.

In this step we, halved the 32 to get the 34 and the −916 comes from need to cancel the extra 916 that is created from multiplying out

(x−34)2=(x−34)(x−34)=x2−34x+916.

Finally we simplify to get 2[(x−34)2−916]−6=2(x−34)2−1816−6=2(x−34)2−578.

We have now finished completing the square and know that

2x2−3x−6=2(x−34)2−578.

Onto actually solving 2x2−3x−6=0. We use our work from above because it doesn't factorise neatly so

2x2−3x−6=02(x−34)2−578=02(x−34)2=578(x−34)2=5716

Now we need to get rid of the 2 by taking the square root, remembering that we're now expecting to use both the postive and negative roots.

(x−34)2=5716x−34=±√5716x=±√5716+34

Which gives the correct answer but we could simplify things - first notice that we can take the square root of 16 on the denominator and then we can make things look a little nicer.

x=±√5716+34=±√574+34=±√57+34

So our final answers are

x=√57+34orx=−√57+34