Express 2x^2-20x+53 in the form 2(x-p)^2+q , where p and q are integers. not sure how to do this..

i factorised it and got this

2(x^2-10x)+53 plz help thx

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- May 16th 2011, 02:32 AMmike789Express 2x^2-28x+53
Express 2x^2-20x+53 in the form 2(x-p)^2+q , where p and q are integers. not sure how to do this..

i factorised it and got this

2(x^2-10x)+53 plz help thx - May 16th 2011, 02:45 AMCaptainBlack
- May 16th 2011, 04:03 AMTheCoffeeMachine
- May 16th 2011, 04:07 AMjgv115
This question is essentially asking you to express that quadratic in "turning point form". This is done by completing the square.

Alternatively, you can "compare coefficients" described in CaptainBlack's post - May 16th 2011, 04:19 AMRomek
$\displaystyle 2x^2 - 20x + 53$

$\displaystyle \equiv 2(x^2 - 10x) + 53$ // Take the 2 outside of the brackets

$\displaystyle \equiv 2((x-5)^2 - 25) + 53$ // Complete the square within the brackets

$\displaystyle \equiv 2(x-5)^2 - 50 + 53$ // Take the -25 outside the brackets

$\displaystyle \equiv 2(x-5)^2 + 3$ // Gather terms - May 16th 2011, 04:44 AMCaptainBlack
What I posted is completing the square, but done from first principles rather than remembering the algorithm, which I don't as it is a waste of brain space (same goes for a number of other things taught in common algebra and starting calculus, they are quicker to re-derive than to remember).

CB - May 16th 2011, 05:08 AMAckbeet
- May 16th 2011, 05:27 AMTheCoffeeMachine