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

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

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

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

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

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

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

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

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

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

But $\displaystyle k = a$, so:

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

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

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

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

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

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

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

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

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

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

But $\displaystyle k = a$, so:

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

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

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

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