Deriving -b forumla, Help please

Username99

I have been trying to work my way through the derivation that I have included as an attachment.

My problem lies with the initial steps.

Step 1:
'a' is factored out and we get left with [ a ][ x^2 + b/a(x) + c/a ] = 0. This is fine.

Step 2:

Are we trying to compare the equation from 'step 1' with the expanded equation of ( x+y )^2 which is x^2 + 2xy + y^2 ?

If so, I can kind of see where the b/2a comes from: Am I correct in saying, if comparing the equation x^2 + 2yx + y^2 to our equation in 'step 1' we need y not 2y to be the co-efficient of the 'x' term, therefore 2yx/2 = 1yx and 1yx is equal to b/2a(x)?

Also he now has made ( x + b/2a )^2, I get that this is the same as (x+y)^2 but where does it come from, is there a step missing between 'step 1' and 'step 2', how was this change performed?

He then takes b^2/4a^2 from this. Again I have a faint idea here, I understand that b^2/ 4a^2 equates to y^2, and if taking away to get rid of y^2 could I rewrite 'step 2' in terms of x and y as:

a[ ( x + y )^2 - y^2 + c/a ] = 0

And if so what does c/a equate to?

From step 3 - step 9 I understand the algebra fully. But I am really lost on how/ why step 1 goes to step 2. Sorry if I have worded my questions terribly, I am a novice.

Any help greatly appreciated.

Yakuza

The jump from step 1 to 2 is that [latex]\frac{b^2}{4a^2}[/latex] have both been added and subtracted to the equation to complete the square - make the square in the format of (x+y)²

A slighly different (and easier to follow), but equivalent explanation is here:
http://www.mathsisfun.com/algebra/quadratic-equation-derivation.html

A somewhat more formal one is here:
http://en.wikipedia.org/wiki/Quadratic_equation#Derivation_of_the_quadratic_formula

Username99

Thanks Yazuka,

I found the Maths is Fun link very helpful, I have no problem with it now.

Just a quick question; Is there any particular reason why a person would solve it
the same way it was solved in my attachment as apposed to the 'Maths is fun' way?

Yakuza

Only if they want to make the derivation a bit more obtuse, to be honest. It's not the clearest on how to get from step 1 to 2. The rest are fairly straightforward though.

MathsManiac

I think the main thing that helps with understanding this is to do plenty of practice with solving quadratics by completing the square first. Once you're a dab hand at this and understand it, the derivation of the formula is a doddle.

I think that completing the square is very useful, as it allows you to find the turning point of the graph (max/min point) by inspection. In certain cases, it's also the quickest way to solve a quadratic, even when the roots are irrational or complex.

Everyone has their own favourite strategy for solving quadratic equations, but here's mine:

1. Fairly obvious factors: -> factorise
2. Else: x^2 co-eff is 1, and x co-eff is even: -> complete the square
3. Else -> formula.

Username99

Thanks i'll take your advice and practice the hell out of it. I don't know how I have passed over completing the square until now. Thanks