How do you show that in the quadratic function: ax^2 + bx + c with 2 roots x1 and x2 that:

1) x1 + x2 = -b/a

and

2) x1*x2 = c/a

?

2. Originally Posted by JQ2009
How do you show that in the quadratic function: ax^2 + bx + c with 2 roots x1 and x2 that:

1) x1 + x2 = -b/a

and

2) x1*x2 = c/a

?
Use the fact that $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$. What happens when you add up the two possible answers?

3. Originally Posted by JQ2009
$x_{1}+x_{2}= -\frac{b}{a}$
Let $x_{1} = \frac{-b+\sqrt{b^2-4ac}}{2a}$

and $x_{2} = \frac{-b-\sqrt{b^2-4ac}}{2a}$

Then calculate $x_{1}+x_{2}= \left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)+\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)$

Alternatively:
Spoiler:
If $x_{1}$ and $x_{2}$ are the roots of the equation $f(x) = ax^2+bx+c$, then the equation must be of the form

$f(x) = k(x-x_{1})(x-x_{2})$ for some constant $k$.

Therefore, we have:
$k(x-x_{1})(k-x_{2}) \equiv ax^2+bx+c$

$\implies k(x^2-(x_{1}+x_{2})x+x_{1}x_{2} \equiv ax^2+bx+c$

$kx^2-(x_{1}+x_{2})kx+x_{1}x_{2} \equiv ax^2+bx+c
$

Equating the coefficients of x gives:
$-k(x_{1}+x_{2}) = b \implies x_{1}+x_{2} = -\frac{b}{k}$

But $k = a$, so:

$\boxed{x_{1}+x_{2} = -\frac{b}{a}}$

Originally Posted by JQ2009
$\left(x_{1}\right)\left(x_{2}\right) = \frac{c}{a}$
Let $x_{1} = \frac{-b+\sqrt{b^2-4ac}}{2a}$

and $x_{2} = \frac{-b-\sqrt{b^2-4ac}}{2a}$

Then calculate $\left(x_{1}\right)\left(x_{2}\right) = \left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)$

Alternatively:
Spoiler:
If $x_{1}$ and $x_{2}$ are the roots of the equation $f(x) = ax^2+bx+c$, then the equation must be of the form

$f(x) = k(x-x_{1})(x-x_{2})$ for some constant $k$.

Therefore, we have:
$k(x-x_{1})(k-x_{2}) \equiv ax^2+bx+c$

$\implies k(x^2-(x_{1}+x_{2})x+x_{1}x_{2} \equiv ax^2+bx+c$

$kx^2-(x_{1}+x_{2})kx+x_{1}x_{2} \equiv ax^2+bx+c
$

Equating the constants gives:
$kx_{1}x_{2} = c \implies x_{1}x_{2} = \frac{c}{k}$

But $k = a$, so:

$\boxed{x_{1}x_{2} = \frac{c}{a}}$

4. $ax^2+bx+c = (X-X_1)(X-X_2)$

$X^2-(X_1 + X_2)x + X_1X_2$

then you can plug in the values for a, b and c