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

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justlooking for someone else

Posts: 77
Registered: 12/10/04
Re: quadratic formula
Posted: Oct 18, 2012 8:11 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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.

s




On Wed, 17 Oct 2012 19:22:35 -0700, William Elliot <marsh@panix.com>
wrote:

>readability On Wed, 17 Oct 2012, stoneboy wrote:
>

>> I have a little problem that I have not been able to get to the form
>> required. I may need some help.
>>
>> here is the problem:
>>
>> if p(x) = ax^2+bx+c is a second degree polynomial, then prove
>>
>> p(x) = a(x+b/2a)^2 - ((b^2 - 4ac)/2a)

>
>Have you writtent he problem correctly?
>No proof is possible. Consider the case b = 0, c /= 0.
>
>If ax^2 + c = p(x) = ax^2 + 4ac/2a = ax^2 + 2c, then c = 0.
>

>> Here I know the roots of the equation for the quadratic equation is
>> x=(-b+sqrt(b^2-4ac))/2a

>
>Pleaseusespacesforreadibility!
>

>> and the negative of the radicand and also
>> x=-b/2a is the vertex
>>

>
>

>> Which root is substituted to get the above form. Or in other words,
>> how do I determine whether I use the negative or the positive radicand
>> in the above equation to get the form shown above. I have simplified
>> it all known ways but cannot seem to get it to that form. Am I
>> barking up the wrong tree here?
>>
>> thanks
>>
>> s
>>




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