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Replies: 5   Last Post: Oct 21, 2012 11:06 AM

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 Frederick Williams Posts: 2,164 Registered: 10/4/10
Posted: Oct 18, 2012 9:13 AM

stoneboy wrote:
>
> my apologies. I should have used spaces for clarity. add to that I
> made a typo. darn it. sorry to waste your time folks.
>
> if p(x) = ax^2 + bx + c is a second degree polynomial, then
> prove
>
>
> p(x) = a(x+b/2a)^2 - ( (b^2 - 4ac) / 4a )
>
> funny how they want me to derive the latter from the former. Guess I
> am a little lost.

It seems to be an odd question. To prove that the second expression for
p(x) is equal to the first, just multiply out the RHS of the second
expression. Far more sensible is to ask: given 'ax^2 + bx + c' how does
one get 'a(x+b/2a)^2 - ((b^2 - 4ac)/(4a))'? The answer to that is 'by
completing the square'. See
http://en.wikipedia.org/wiki/Completing_the_square, for example.

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Till shade and silence waken up as one,
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