Challenge Problem

• Sep 26th 2009, 01:21 PM
fizz
Challenge Problem
Given: $ax^2+bx+c=0$ and $(x-p)(x-q)=0$,

Prove: $p+q=-b/a$ and $pq=c/a$
• Sep 26th 2009, 01:26 PM
artvandalay11
Apply the quadratic formula to $ax^2+bx+c=0$

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 $-\frac{b}{a}$ and multiply them together to get $\frac{c}{a}$