How do I find the sum and product of the roots of this equation?
The equation is $\displaystyle 7x^2-4x+2=0$.
Well, the direct way is to solve the equation and find the sum and the product!
To show how to do this without actually solving:
$\displaystyle x_{\pm} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
So
$\displaystyle x_{+} + x_{-} = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-b}{2a} + \frac{-b}{2a} = -\frac{b}{a}$
and
$\displaystyle x_{+} \cdot x_{-} = \left ( \frac{-b + \sqrt{b^2 - 4ac}}{2a} \right ) \left ( \frac{-b - \sqrt{b^2 - 4ac}}{2a} \right ) $
Recall that $\displaystyle (p + q)(p - q) = p^2 - q^2$, so
$\displaystyle x_{+} \cdot x_{-} = \left ( \frac{-b}{2a} \right ) ^2 - \left ( \frac{\sqrt{b^2 - 4ac}}{2a} \right ) ^2$
$\displaystyle x_{+} \cdot x_{-} = \frac{b^2}{4a^2} - \frac{b^2 - 4ac}{4a^2} = \frac{c}{a}$
-Dan
Hello,
you can use the theorem of Vieta(?) (that's the name I know for this theorem)
if you have a quadratic equation:
$\displaystyle ax^2+bx+c=0~\Longrightarrow~x^2+\frac{b}{a}x+\frac {c}{a}=0$ with the solutions $\displaystyle x_1 \text{ and } x_2$ then the theorem says:
$\displaystyle x_1 \cdot x_2 = \frac{c}{a}$ and
$\displaystyle x_1 + x_2 = -\frac{b}{a}$