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# Proving Ax^2 + Bx + C watch

1. Hi, i was looking at two revision guides last nite- one was Letts and i cant remember the other. They were showing how to prove the quadratic equation. One of them started by squaring the root of the equation (which was slightly boggling to follow) but the other one did it in such an easy way. The thing was, it was as if they were making it up as they went along. Firstly they multiplied both sides by 4a, then added b^2 to complete the square and subtract it to keep it level and worked from there, but this seemed so much easier than the other method.

My only question is why did they multiply by 4a at the start? I mean, what is the mathematical reason.....
2. Can you write up the full proof for us? So that we can see what this original equation is? Sorry but your original post was a bit fuzzy.
3. Um, the derivation I'm familiar with is:

ax² + bx + c = 0
x² + (b/a)x + (c/a) = 0

Rearrange to give

[x + b/2a]² - (b²/4a²) + (c/a) = 0
[x + b/2a]² = (b²/4a²) - (c/a)

You multiply (c/a), top and bottom, by 4a to get

(b² - 4ac)/4a² on the right hand side

and the rest of the derivation is simple algebra

That's where the 4a comes in anyway
4. hey squishy, i'll go one better and scan it for you- hang on a few mins
5. Attached Images

6. it isnt that i dont understand what they've done- i just dont understand why that particular route/method (it was late last nite i was looking! lol)
7. So you get 4ac, well that's as far as I can see, if you get 4ac somewhere then it's going to make things easier to rearange.
8. Just a question:

Imagine

4x^2+6x+4 which =

(4x+3)^2-5 (3 squared equals 9, so you minus 5 to keep it equal)

The part in the proof that i dont understand, is (with the version above) when you square 3, you then have to minus 5 to end up with +4. Where does this happen in (for example) squishy's version? Thats what puzzles me a little.
9. It happens with:

"x² + (b/a)x + (c/a) = 0

Rearrange to give

[x + b/2a]² - (b²/4a²) + (c/a) = 0"

you know completing the square? you take something which isn't a perfect square, say:

x^2 + 6x

And as a rule, half the coefficient of x, add this number to x and square the brackets. But that's not all, because:

(x + 3)^2 = x^2 + 6x +9

THere's an extra nine that you've added by doing this procedure that you don't want. SO by taking nine from the whole thing, you get:

x^2 + 6x = (x + 3)^2 - 9

which, btw (throgh difference of 2 squares)

(x + 3)^2 - 9 = [(x+3) + 3][(x+3) - 3] = x(x+6) = x^2 + 6x
10. yeap, just to make a complete square.
You can also multiply both side by a to get
(ax)^2 + (ax)*b + ac = 0.
u knew A^2 + 2AB + B^2 = (A + B)^2 .plus then minus (b/2)^2 to get
(ax)^2 + 2*(ax)(b/2) + (b/2)^2 - (b/2)^2 + ac = 0.
(ax + b/2)^2 = (b/2)^2 - ac.

As squishy did, multiplying by 4a is just another way to make a complete square.
11. yea thats ok, i understood all that- i was just tryin to see it in squishys formula ax^2+bx+c (replacing letters for numbers).

So can someone tell me what is the theory behind completing the square, is it just an alternative for factorising and the quadratic equation? I like to know what and why for everything i do, i'd rather know why im doing somewthing- than to just carry it out like a robot.
12. (Original post by faa)
yea thats ok, i understood all that- i was just tryin to see it in squishys formula ax^2+bx+c (replacing letters for numbers).

So can someone tell me what is the theory behind completing the square, is it just an alternative for factorising and the quadratic equation? I like to know what and why for everything i do, i'd rather know why im doing somewthing- than to just carry it out like a robot.
You do it to solve a quadratic, if it doesn't factorise. Eg.

x^2 + 6x + 1 = 0

This doesn't factorise, so you're nore gonna have nice roots. You COULD use the quadratic formula, but it's more stylish to do this (complete the square):

x^2 + 6x + 1 = 0
(x + 3)^2 + 1 - 9 = 0
(x + 3)^2 = 8
(x + 3) = +/- sqrt8 = +/- 2.sqrt2
x = +/- 2.sqrt2 - 3

So after doing this, the correct factorisation was:

x^2 + 6x + 1 = (x + 2.sqrt2 - 3)(x - 2.sqrt2 - 3), but we don't need this now that we have the answers.

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