Completing the square

Ive just learnt about completing the square, and I understand how we do it. But I don't get why we do it? can someone explain this to me? I have a question that Ive answered and was wondering if you could refer to this write in the form of (x+p)squared + q xsquared +8x+15 I did (x+4)squared-1... thanks :-)

Your answer is correct, and it's useful for plotting points as said above, and at a level it's used to find minimum points of graphs.

If we didn't know how to complete the swquare, we wouldn't be able to get the quadratic formula basically.

I found a way of deriving it without completing the square, but it is ridiculously unintuitive. I much prefer the conpleting the square method.

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OK...
$[br]ax^{2} + bx + c =0[br]4a^{2}x^{2} + 4abx + 4ac = 0[br]4a^{2}x^{2} + 4abx + 4ac + b^{2} = b^{2}[br]4a^{2}x^{2} + 4abx + b^{2} = b^{2} - 4ac[br]2ax + b = \pm\sqrt{b^{2}-4ac}[br]2ax = -b\pm\sqrt{b^{2}-4ac}[br]x = \dfrac {-b\pm\sqrt{b^{2}-4ac}}{2a}[br]$
Replace + at the end with plus/minus, don't know how to do it in latex!


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Forgive me if I'm being stupid, but how did you get from step 4 to step 5? (I don't know how to use latex) I don't understand how you got 2ax + b = square root of 4a^2x^2 + 4abx + b^2. I know this is true, simply by plugging in some made up values, but how did you get there via algebra?

I'm sure there's a 'proper' way but I just did it by inspection. In that situation you know that 4a^2x^2 when started is equal to 2ax, and b^2 sqrted is obviously b, so if you test (2ax+b)^2 you can see that it works. In all truth to get this I worked backwards from the quadratic formula, and it is appreciably easier that way


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