Formula for Sum and Product of Roots

Sum of Roots:    Product of Roots:
$\displaystyle \alpha +\beta =-\frac{b}{a}$       $\displaystyle \alpha \times \beta =\frac{c}{a}$

From the Quadratic Equation:  a = 3, b = 2p and c = -p

a) 2 equal roots.

Let the roots be α & α since the roots are equal . Do note that we can also let the roots be β and β.

Sum of Roots:

$\displaystyle \alpha +\alpha =-\frac{{2p}}{3}$
$\displaystyle 2\alpha =-\frac{{2p}}{3}$
$\displaystyle \alpha =-\frac{p}{3}\quad \to (1)$

Product of Roots:

$\displaystyle \alpha \times \alpha =\frac{p}{3}$

$\displaystyle {{\alpha }^{2}}=\frac{p}{3}\quad \to (2)$

Substitute Equation (1) into (2)

$\displaystyle {{\left( {-\frac{p}{3}} \right)}^{2}}=\frac{p}{3}$

$\displaystyle \frac{{{{p}^{2}}}}{9}=\frac{p}{3}$

$\displaystyle 3{{p}^{2}}=9p$

$\displaystyle 3{{p}^{2}}-9p=0$

$\displaystyle 3p(p-3)=0$

$\displaystyle 3p=0\ or\ p-3=0$

$\displaystyle p=0\ or\ p=3$

b) 1 roots is a reciprocal of the other

Let the roots be α & 1/α since the roots are equal . Do note that we can also let the roots be β and 1/β.

Sum of Roots:

$\displaystyle \alpha +\frac{1}{\alpha }=-\frac{{2p}}{3}$

$\displaystyle \frac{{{{\alpha }^{2}}}}{\alpha }+\frac{1}{\alpha }=-\frac{{2p}}{3}$

$\displaystyle \frac{{{{\alpha }^{2}}+1}}{\alpha }=-\frac{{2p}}{3}\quad$

There are two unknown variables α and p, we cannot find the value of p, so we try to find for p using product of roots.

Product of Roots:

$\displaystyle \alpha \times \frac{1}{\alpha }=\frac{p}{3}$

$\displaystyle 1=\frac{p}{3}\quad$

$\displaystyle p=3\$

Students assume that the roots of the equation are always α and β. I purposely posted this solution to show all of you that the roots does not always have to be so.