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

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 justlooking for someone else Posts: 77 Registered: 12/10/04
Posted: Oct 21, 2012 11:06 AM

Thank you for your explanation, Frederick. You are correct in how I
should have interpreted that question in the first place. I re-read
the question and came to the conclusion that you did. I reviewed the
wiki on completing the square and that was the clue to the required
solution. Appreciate your time and insight into that problem. Thanks

s

On Thu, 18 Oct 2012 14:13:38 +0100, Frederick Williams
<freddywilliams@btinternet.com> wrote:

>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.