Express 2x^2-28x+53

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May 16th 2011, 02:32 AM
mike789

Express 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 AM
CaptainBlack

Quote:

Originally Posted by mike789

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

If:

$2(x-p)^2+q=2x^2-4px+2p^2+q=2x^2-20x+53$

then $4p=20$ , so $p=5$ , and $2p^2+q=53$ so $q= .... ?$

CB
May 16th 2011, 04:03 AM
TheCoffeeMachine

Quote:

Originally Posted by mike789

i factorised it and got this
2(x^2-10x)+53 plz help thx

Good start! Now consider (x-5)² = x²-10x+25 which gives x²-10x = (x-5)²-25.
May 16th 2011, 04:07 AM
jgv115

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 AM
Romek

$2x^2 - 20x + 53$
$\equiv 2(x^2 - 10x) + 53$ // Take the 2 outside of the brackets
$\equiv 2((x-5)^2 - 25) + 53$ // Complete the square within the brackets
$\equiv 2(x-5)^2 - 50 + 53$ // Take the -25 outside the brackets
$\equiv 2(x-5)^2 + 3$ // Gather terms
May 16th 2011, 04:44 AM
CaptainBlack

Quote:

Originally Posted by jgv115

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

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 AM
Ackbeet

Quote:

Originally Posted by CaptainBlack

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

Yes, yes. To CB you listen!
May 16th 2011, 05:27 AM
TheCoffeeMachine

Quote:

Originally Posted by CaptainBlack

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).

If you are referring to remembering the formula for the coordinates of the turning point, I agree. But what I was hinting at post #3 doesn't really require remembering stuff at all.