quadratic function

JQ2009 · #1 $Old$ November 2nd, 2009, 09:02 AM

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

?

Jameson · #2 $Old$ November 2nd, 2009, 09:13 AM

Quote:

Originally Posted by JQ2009 $View Post$

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?

I4talent · #3 $Old$ November 2nd, 2009, 10:20 AM

Quote:

Originally Posted by JQ2009 $View Post$

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

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:

Quote:

Originally Posted by JQ2009 $View Post$

$\left(x_{1}\right)\left(x_{2}\right) = \frac{c}{a}$

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:

sammy28 · #4 $Old$ November 2nd, 2009, 10:48 AM

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

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

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