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Re: quadratic formula
Posted:
Oct 18, 2012 9:13 AM
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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.
--
Where are the songs of Summer?--With the sun,
Oping the dusky eyelids of the south,
Till shade and silence waken up as one,
And morning sings with a warm odorous mouth.
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