completing the square

**GAdams** · September 27th 2007, 10:57 AM

The second last sentence says that we can only use completing the square when the x coefficient is 1. But the equation has an x coefficient of 6??

**earboth** · September 27th 2007, 11:03 AM

Originally Posted by GAdams

The second last sentence says that we can only use completing the square when the x coefficient is 1. But the equation has an x coefficient of 6??

Hello,

that's a typo. It should say: The technique is only valid if 1 is the factor of $x^2$

**GAdams** · September 27th 2007, 11:09 AM

OK. Thanks.

So if x^2 has a coefficient of more than one, then we can't use this method?

It's only that I have an equation that I was told could be done via this method. Hmmm.

2x^2 + 3x - 2

**GAdams** · September 27th 2007, 11:14 AM

Does anybody know a weblink that explains completing the square with explanations of why they do what they do or something?

I just can't get my head around it.

Thanks.

**DivideBy0** · September 27th 2007, 11:15 AM

Nah, you still can, you just need to factor out the coefficient of $x^2$ so you get 1 as the coefficient.

$2x^2 + 3x - 2$

$=2(x^2+\frac{3}{2}x-1)$

Continue normally from here.

Here are some links you might like to try:
Completing the Square
Completing the Square: Solving Quadratic Equations

**GAdams** · September 28th 2007, 12:36 AM

I tried an easier one first. Does this look ok?

**GAdams** · September 28th 2007, 01:01 AM

This is as far as i got...I seem to complicate it more if I try to simplify further

**DivideBy0** · September 28th 2007, 03:23 AM

You don't need to get all confused with $a,b,c$ when completing the square. $a,b,c$ are just used in the theory of completing the square. They are good for when you want to understand what's actually going on, but don't worry about them when you're actually doing a problem.
When I was learning how to factorise with completing the square I repeated this mantra:
"Add and subtract the square of half the coefficient of the x term". That's practically all there is to it.

Originally Posted by GAdams

I tried an easier one first. Does this look ok?

$x^2+6x-7=0$

$x^2+6x+\left(\frac{6}{2} \right)^2-\left(\frac{6}{2} \right)^2-7=0$

$(x+3)^2-\left(\frac{6}{2}\right)^2-7=0$

$(x+3)^2-16=0$

$(x+3)^2=16$

$x+3=\pm 4$

$x=-3 \pm 4$

$x=-7$ or $x=1$

(You could probably do this faster with trial and error factorisation)

Originally Posted by GAdams

This is as far as i got...I seem to complicate it more if I try to simplify further

For this, I'll just factorise it,

$2x^2+3x-2$

$=2\left(x^2+\frac{3}{2}x-1\right)$

$=2\left(x^2+\frac{3}{2}x+\left(\frac{3}{4} \right)^2-\left(\frac{3}{4} \right)^2-1\right)$

$=2\left(\left(x+\frac{3}{4}\right)^2-\frac{25}{16}\right)$

Since this is the same as

$=2\left(\left(x+\frac{3}{4}\right)^2-\left(\sqrt{\frac{25}{16}}\right)^2\right)$

Now you can use difference of perfect squares $a^2-b^2=(a-b)(a+b)$ to factorise this further:

$=2\left(x+\frac{3}{4}+\sqrt{\frac{25}{16}}\right)\ left(x+\frac{3}{4}-\sqrt{\frac{25}{16}}\right)$

$=2\left(x+\frac{3}{4}+\frac{5}{4}\right)\left(x+\f rac{3}{4}-\frac{5}{4}\right)$

$=2(x+2)(x-\frac{1}{2})$

**Krizalid** · September 28th 2007, 07:17 AM

Another attempt

$2x^2+3x-2=2x^2+(4x-x)-2=(2x^2+4x)-(x+2)$

Now

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

**GAdams** · September 28th 2007, 07:40 AM

OK. I got the first one. Thanks.

The second one is still confusing me, I can't see how you got the 3/4.

Here's what I tried. I have gone wrong somewhere....not sure where:

**earboth** · September 28th 2007, 08:04 AM

Originally Posted by GAdams

The second one is still confusing me, I can't see how you got the 3/4.

Here's what I tried. I have gone wrong somewhere....not sure where:

Hello,

If you've got an equation and you want to use your method then you have to divide first the complete equation by the coefficient of the x². (Remember: The coefficient of x² must be 1):

$2x^2+3x-2 = 0~\iff~x^2+\frac32 x-1=0$

Now use your method:

$x^2+\frac32 x + \left(\frac34 \right)^2-\left(\frac34 \right)^2-1=0~\iff ~\left(x+\frac34 \right)^2-\left(\frac9{16} + \frac{16}{16}\right)=0$

You've got a difference of squares which can be factored easily:
$<br /> \left(x+\frac34 \right)^2-\left(\frac9{16} + \frac{16}{16}\right)=0~\iff~\left(x+\frac34 +\frac54\right)\left(x+\frac34 -\frac54\right)=0~$ $\iff~(x+2)(x-\frac12)=0$

I'll leave the rest for you.

**topsquark** · September 28th 2007, 10:30 AM

GAdams: You seem to be missing the point of this method. What you are trying to do here is turn the expression
$x^2 + bx$
into
$x^2 + bx + \frac{b^2}{4} - \frac{b^2}{4}$

Why are you doing this? Because looking at just the first three terms:
$x^2 + bx + \frac{b^2}{4} = \left ( x + \frac{b}{2} \right )^2$

This process gives us an equation that is much easier to solve than the original.

As an example, solve
$2x^2 + 7x - 11 = 0$

First group the first two terms:
$(2x^2 + 7x) - 11 = 0$

Now factor the coefficient of the $x^2$ term:
$2 \left ( x^2 + \frac{7}{2} x \right ) - 11 = 0$

Now let's complete the square. The coefficient of the linear term is $\frac{7}{2}$ . We want to take half of that and square it. But we want to do so in such a way that we aren't adding anything to the LHS of the equation. So we're going to add a term and subtract the same to give a net change of 0:
$2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 - \left ( \frac{7}{4} \right )^2 \right ) - 11 = 0$

Now rearrange a bit:
$2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 \right ) - 2 \left ( \frac{7}{4} \right )^2 - 11 = 0$

Now the first three terms is a perfect square:
$2 \left ( x + \frac{7}{4} \right ) ^2 - \frac{49}{8} - 11 = 0$

Now to neaten a few things up:
$2 \left ( x + \frac{7}{4} \right ) ^2 - \frac{137}{8} = 0$

$2 \left ( x + \frac{7}{4} \right ) ^2 = \frac{137}{8}$

$\left ( x + \frac{7}{4} \right ) ^2 = \frac{137}{16}$

$x + \frac{7}{4} = \pm \sqrt{\frac{137}{16}}$

$x = -\frac{7}{4} \pm \frac{\sqrt{137}}{4}$

-Dan

**GAdams** · September 29th 2007, 02:35 AM

Originally Posted by topsquark

GAdams: You seem to be missing the point of this method. What you are trying to do here is turn the expression
$x^2 + bx$
into
$x^2 + bx + \frac{b^2}{4} - \frac{b^2}{4}$

Why are you doing this? Because looking at just the first three terms:
$x^2 + bx + \frac{b^2}{4} = \left ( x + \frac{b}{2} \right )^2$

This process gives us an equation that is much easier to solve than the original.

As an example, solve
$2x^2 + 7x - 11 = 0$

First group the first two terms:
$(2x^2 + 7x) - 11 = 0$

Now factor the coefficient of the $x^2$ term:
$2 \left ( x^2 + \frac{7}{2} x \right ) - 11 = 0$

Now let's complete the square. The coefficient of the linear term is $\frac{7}{2}$ . We want to take half of that and square it. But we want to do so in such a way that we aren't adding anything to the LHS of the equation. So we're going to add a term and subtract the same to give a net change of 0:
$2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{2} \right )^2 - \left ( \frac{7}{2} \right )^2 \right ) - 11 = 0$

-Dan

OK. You said 'take half of that and square it. But you haven't halved it. It's still 7/2 -you just squared it.

Did you mean half the original coefficient of 7?

**topsquark** · September 29th 2007, 03:13 AM

Originally Posted by GAdams

OK. You said 'take half of that and square it. But you haven't halved it. It's still 7/2 -you just squared it.

Did you mean half the original coefficient of 7?

Nope. I did a typo.

I have fixed it in my original post. Sorry about that. (But hey! You were able to follow it and spotted the mistake!

)

-Dan

**GAdams** · September 29th 2007, 06:44 AM

Originally Posted by topsquark

So we're going to add a term and subtract the same to give a net change of 0:
$2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 - \left ( \frac{7}{4} \right )^2 \right ) - 11 = 0$

Now rearrange a bit:
$2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 \right ) - 2 \left ( \frac{7}{4} \right )^2 - 11 = 0$

Now the first three terms is a perfect square:
$2 \left ( x + \frac{7}{4} \right ) ^2 - \frac{49}{8} - 11 = 0$

-Dan

I can't see how you got teh step after you re-arranged.

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