quadratic function

Quote:

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

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

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

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

But $k = a$ , so:

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

Quote:

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

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

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

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

But $k = a$ , so:

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