# Thread: I REALLY need help with this quadratic proof.. Would appriciate it.

1. ## I REALLY need help with this quadratic proof.. Would appriciate it.

QUESTION: Suppose that r and s are distinct real solutions of the quadratic equation ax^2 + bx + c = 0. In terms of a,b,c, obtain forumlas for r + s and rs.

SOLUTION?
so what i know the answer is r+s= -b/a and rs=c/a but im confused as to why.

SO i started this...
suppose r=x and s=y:
(x+y)= M ( where M is an element of the real #)
(xy)=T (where M is an element of the real #)

so then x(M-x)=T
x^2-Mx+T=0
ax^2-aMx+aT=0

assuming aM =b and aT=c we get out original quadratic equation.

Now im stuck.. i tried solving for r+s(aMx) using what i have but i get something different. Please help =/

2. If you know the Quadratic Formula, then the two solutions are $\displaystyle \displaystyle \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ and $\displaystyle \displaystyle \frac{-b - \sqrt{b^2 - 4ac}}{2a}$.

What do you get when you add them? What do you get when you multiply them?

3. Originally Posted by kensington
QUESTION: Suppose that r and s are distinct real solutions of the quadratic equation ax^2 + bx + c = 0. In terms of a,b,c, obtain forumlas for r + s and rs.
First of all, what are the solutions for $\displaystyle \displaystyle ax^2 + bx + c = 0 \implies x = \frac{-b\pm\dots }{2a}$ ?

4. Originally Posted by Prove It
If you know the Quadratic Formula, then the two solutions are $\displaystyle \displaystyle \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ and $\displaystyle \displaystyle \frac{-b - \sqrt{b^2 - 4ac}}{2a}$.

What do you get when you add them? What do you get when you multiply them?
Thank you sooo much!

5. Originally Posted by kensington
QUESTION: Suppose that r and s are distinct real solutions of the quadratic equation ax^2 + bx + c = 0. In terms of a,b,c, obtain forumlas for r + s and rs.

SOLUTION?
so what i know the answer is r+s= -b/a and rs=c/a but im confused as to why.

SO i started this...
suppose r=x and s=y:
(x+y)= M ( where M is an element of the real #)
(xy)=T (where M is an element of the real #)

so then x(M-x)=T
x^2-Mx+T=0
ax^2-aMx+aT=0

assuming aM =b and aT=c we get out original quadratic equation.

Now im stuck.. i tried solving for r+s(aMx) using what i have but i get something different. Please help =/
$\displaystyle a(x-r)(x-s)=ax^2+bx+c$

Expand the left hand side:

$\displaystyle ax^2-a(r+s)x+a(rs)=ax^2+bx+c$

Now equate coefficients of like powers of $\displaystyle$$x$.

CB