Given: $\displaystyle ax^2+bx+c=0$ and $\displaystyle (x-p)(x-q)=0$,

Prove: $\displaystyle p+q=-b/a$ and $\displaystyle pq=c/a$

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- Sep 26th 2009, 01:21 PMfizzChallenge Problem
Given: $\displaystyle ax^2+bx+c=0$ and $\displaystyle (x-p)(x-q)=0$,

Prove: $\displaystyle p+q=-b/a$ and $\displaystyle pq=c/a$ - Sep 26th 2009, 01:26 PMartvandalay11
Apply the quadratic formula to $\displaystyle ax^2+bx+c=0$

The two answers you get are the roots right?

So let's say you get p,q to be the roots (which is what the question is saying) then you can write the original formula as (x-p)(x-q)=0

When you get the two roots from the quadratic formula, add them together and you'll get $\displaystyle -\frac{b}{a}$ and multiply them together to get $\displaystyle \frac{c}{a}$