1. ## Completing the square.

Well, I am working on my quadratics at the moment, and one of my main problems with it is completing the square. I have a basic idea on how to do it, but could anyone explain it from start to finish in detail. That would be great, or if you just have a link to a good site that'd be great. Cheers

2. Originally Posted by tumbleweed
Well, I am working on my quadratics at the moment, and one of my main problems with it is completing the square. I have a basic idea on how to do it, but could anyone explain it from start to finish in detail. That would be great, or if you just have a link to a good site that'd be great. Cheers
Here's an example.

Complete the square: $y^2-4y-3=x$

When completing the square, you have to think a little bit about what the left side of this equation looks like. You need to add something to the left side to make it a perfect square. A little bit of thought leads us to the conclusion that you need to have $y^2-4y+4$ on the left side because that can be factored down to $(y-2)^2$. We want to take all the constants on the left side over to the other side to give us a blank slate to work with on the left side.

$y^2-4y-3+3=x+3$
$y^2-4y=x+3$
Now, add 4 to both sides to "complete the square":
$y^2-4y+4=x+3+4$
$y^2-4y+4=x+7$
Factor.
$(y-2)^2=x+7$

The trickiest part is knowing what constant has to be added to the left side in order to "complete the square". You can figure it out by recognizing that the expansion of $(y-c)^2$ is $y^2-2cy+c^2$. Since you know the middle term (2cy) corresponds to -4y, you know that $2cy=-4y$. Cancel the y's and divide both sides by 2. $c=-2$. Since the third term in the expansion is $c^2$, you know it must be +4. That's how you know it has to be $y^2-4y+4$.

That's about all there is to it. I hope this was clear enough...

Pulsar

3. Originally Posted by tumbleweed
Well, I am working on my quadratics at the moment, and one of my main problems with it is completing the square. I have a basic idea on how to do it, but could anyone explain it from start to finish in detail. That would be great, or if you just have a link to a good site that'd be great. Cheers
Another way to do it is this:

y^2 - 4y - 3 = x

Now we need to take half of b (4) and square it. Thats how I was taught to do it. This makes sense because if you think about it, in the end it would workout to be a perfect square. So "b" is taken from ax^2 + bx + c = 0. The first step is to always take half of b and square it. [(1/2b)^2] In this question, you get 4. Now we must add and subtract 4 on the left side. We need that 4 in there to get a perfect square, but we cannot randomly add it. Therefore, we add and subtract it, as if we didn't do anything at all, because 4-4 is just 0. Also, we leave the "c" value right where it is and don't include it in the brackets. So we get :

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

Now we can have a perfect square using y^2 - 4y + 4, giving us:

[(y-2)^2 -4)] - 3 = x

Then, we add the -4 and the -3 together giving us:

(y-2)^2 - 7 = x

4. Hello, tumbleweed!

I assume you're solving a quadratic equation.

An additional problem arises if the leading coefficient is not 1 (one).

For example: . $4x^2 - 24x + 11 \:=\:0$

"Move" the constant to the right side: . $4x^2 - 24x \:=\:-11$

Divide through by the leading coefficient: . $x^2 - 6x\qquad \;=\;-\frac{11}{4}$

Complete the square: take half the x-coefficient and square.
. . . . $\frac{1}{2}(\text{-}6) \;=\;-3\quad\Rightarrow\quad(\text{-}3)^2 \;=\;9$

Add to both sides: . $x^2 - 6x\,{\color{blue} +\,9} \;=\;-\frac{11}{4}\,{\color{blue}+\,9}$

And we have: . $(x - 3)^2 \;=\;\frac{25}{4}$

Now we can solve for $x.$

Take the square root of both sides: . $x - 3 \;=\;\pm\sqrt{\frac{25}{4}}$
. . and we have: . $x - 3 \;=\;\pm\frac{5}{2}$

Therefore: . $x \;=\;3\pm\frac{5}{2} \;=\;\begin{Bmatrix}3 + \frac{5}{2} & = & \frac{11}{2} \\ 3 - \frac{5}{2} & = & \frac{1}{2} \end{Bmatrix}$

5. Originally Posted by Soroban
"Move" the constant to the right side: . $4x^2 - 24x \:=\:-11$
Here's another way:

$(2x-6)^2-36=-11\implies|2x-6|=5.$

From here $2x-6=5$ & $6-2x=5,$

which gives Soroban's solutions.