# Thread: How do you complete the square with a negative coefficient?

1. ## How do you complete the square with a negative coefficient?

Hi guys basically I have got a bit of a problem with completing the square...

If I had to complete the square of x2-14x+36=0

I would form the square of (x-7)2

I would then expand the brackets to get x2-14x+49

I would then combine this with the original equation to get x2-14x+49+36=0

I then tidy it up to get x2-14x+85=0

I then take 85 from both sides to leave me with x2-14x=-85

My reaction to that would be to try and remove the 2 by square rooting each side but then I would be rooting a negative? Can someone give me some help by talking me through what to do as opposed to just showing me a page of maths? Sorry but I just don't learn particularly well that way.

Thanks in advance for anyone who helps me, it is appreciated.

2. ## Re: How do you complete the square with a negative coefficient?

Originally Posted by MattA147
Hi guys basically I have got a bit of a problem with completing the square...
If I had to complete the square of x2-14x+36=0
$\displaystyle \\x^2-14x+36=0\\x^2-14x+49+36=49\\(x-7)^2=13$

3. ## Re: How do you complete the square with a negative coefficient?

Originally Posted by MattA147
Hi guys basically I have got a bit of a problem with completing the square...

If I had to complete the square of x2-14x+36=0

I would form the square of (x-7)2

I would then expand the brackets to get x2-14x+49
Yes, that's correct.

I would then combine this with the original equation to get x2-14x+49+36=0
This is where you went wrong. You cannot just add "49" to one side. You CAN, as Plato says, add 49 to both sides of the equation to get
$\displaystyle x^2- 14x+ 49+ 36= 49$
$\displaystyle x^2- 14x+ 49= 49- 36= 13$
$\displaystyle (x- 7)^2= 13$

The other thing you can do is, noting that 36= 49- 13, write this as $\displaystyle x^2- 14x+ 49- 13= 0$, then write it as $\displaystyle (x- 7)^2= 13$.

I then tidy it up to get x2-14x+85=0
It should be clear that this is impossible! If x2- 14x+ 36= 0, then x2- 14x+ 85 cannot also be 0- it is obviously 36 more

I then take 85 from both sides to leave me with x2-14x=-85
And now you have undone the whole point of "completing the square"! The left side, x2- 14x is NOT a "perfect square" so you cannot take the square root.

My reaction to that would be to try and remove the 2 by square rooting each side but then I would be rooting a negative? Can someone give me some help by talking me through what to do as opposed to just showing me a page of maths? Sorry but I just don't learn particularly well that way.

Thanks in advance for anyone who helps me, it is appreciated.
It is, of course, possible that completing the square correctly will give a negative number. For example, if you have x2- 2x+ 4= 0, you would "complete the square" by writing 4 as 1+ 3: x3- 2x+ 1+ 3= (x-1)2+ 3= 0 which is the same as (x- 1)2= -3. Now you could either (depending on the "level" of the course) say "There is no (real number) value of x that satisfies the equation so the equation has no solution" or say that $\displaystyle x- 1= \pm i\sqrt{3}$ so that the two solutions are [tex]x= 1\pm i\sqrt{3}[tex].

4. ## Re: How do you complete the square with a negative coefficient?

Thanks very much guys for highlighting where I went wrong. I appreciate the time, and I will be sure to not make the same mistake/s in the future.

5. ## Re: How do you complete the square with a negative coefficient?

To avoid getting into issue of 85 etc we may proceed as under:
x^2 - 14 x + 36 = 0
step - 1 get rid of constant from LHS. In this case by adding ( -36) on both sides we get
x^2 - 14 x = - 36
Step 2. Now add square of half the coefficient of x on both sides. in this case the coefficient of x is -14, half is -7 square of ( -7) = 49.
Thus we add 49 on both the sides of the equation.
x^2 - 14 x + 49 = 49 - 36
( x-7)^2 = 13