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??

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- Sep 27th 2007, 10:57 AMGAdamscompleting the square
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??

- Sep 27th 2007, 11:03 AMearboth
- Sep 27th 2007, 11:09 AMGAdams
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 - Sep 27th 2007, 11:14 AMGAdams
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. - Sep 27th 2007, 11:15 AMDivideBy0
Nah, you still can, you just need to factor out the coefficient of $\displaystyle x^2$ so you get 1 as the coefficient.

$\displaystyle 2x^2 + 3x - 2$

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

Continue normally from here.

Here are some links you might like to try:

Completing the Square

Completing the Square: Solving Quadratic Equations - Sep 28th 2007, 12:36 AMGAdams
I tried an easier one first. Does this look ok?

- Sep 28th 2007, 01:01 AMGAdams
This is as far as i got...I seem to complicate it more if I try to simplify further

- Sep 28th 2007, 03:23 AMDivideBy0
You don't need to get all confused with $\displaystyle a,b,c$ when completing the square. $\displaystyle a,b,c$ 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.

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

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

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

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

$\displaystyle (x+3)^2=16$

$\displaystyle x+3=\pm 4$

$\displaystyle x=-3 \pm 4$

$\displaystyle x=-7$ or $\displaystyle x=1$

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

For this, I'll just factorise it,

$\displaystyle 2x^2+3x-2$

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

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

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

Since this is the same as

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

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

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

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

$\displaystyle =2(x+2)(x-\frac{1}{2})$ - Sep 28th 2007, 07:17 AMKrizalid
Another attempt

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

Now

$\displaystyle 2x(x+2)-(x+2)=(x+2)(2x-1)$ - Sep 28th 2007, 07:40 AMGAdams
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: - Sep 28th 2007, 08:04 AMearboth
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):

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

Now use your method:

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

You've got a difference of squares which can be factored easily:

$\displaystyle

\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~$$\displaystyle \iff~(x+2)(x-\frac12)=0$

I'll leave the rest for you. - Sep 28th 2007, 10:30 AMtopsquark
GAdams: You seem to be missing the point of this method. What you are trying to do here is turn the expression

$\displaystyle x^2 + bx$

into

$\displaystyle x^2 + bx + \frac{b^2}{4} - \frac{b^2}{4}$

Why are you doing this? Because looking at just the first three terms:

$\displaystyle x^2 + bx + \frac{b^2}{4} = \left ( x + \frac{b}{2} \right )^2$

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

As an example, solve

$\displaystyle 2x^2 + 7x - 11 = 0$

First group the first two terms:

$\displaystyle (2x^2 + 7x) - 11 = 0$

Now factor the coefficient of the $\displaystyle x^2$ term:

$\displaystyle 2 \left ( x^2 + \frac{7}{2} x \right ) - 11 = 0$

Now let's complete the square. The coefficient of the linear term is $\displaystyle \frac{7}{2}$. 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:

$\displaystyle 2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 - \left ( \frac{7}{4} \right )^2 \right ) - 11 = 0$

Now rearrange a bit:

$\displaystyle 2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 \right ) - 2 \left ( \frac{7}{4} \right )^2 - 11 = 0$

Now the first three terms is a perfect square:

$\displaystyle 2 \left ( x + \frac{7}{4} \right ) ^2 - \frac{49}{8} - 11 = 0$

Now to neaten a few things up:

$\displaystyle 2 \left ( x + \frac{7}{4} \right ) ^2 - \frac{137}{8} = 0$

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

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

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

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

-Dan - Sep 29th 2007, 02:35 AMGAdams
- Sep 29th 2007, 03:13 AMtopsquark
- Sep 29th 2007, 06:44 AMGAdams