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??
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
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:
$\displaystyle =2\left(x+\frac{3}{4}+\sqrt{\frac{25}{16}}\right)\ left(x+\frac{3}{4}-\sqrt{\frac{25}{16}}\right)$
$\displaystyle =2\left(x+\frac{3}{4}+\frac{5}{4}\right)\left(x+\f rac{3}{4}-\frac{5}{4}\right)$
$\displaystyle =2(x+2)(x-\frac{1}{2})$
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.
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