Quadratic has Fractions!

First equation, $ax² + bx + c = 0$ with $p$ and $q$ as the roots.

By Vieta's Formula, we have $p + q = -\frac{b}{a} ...(1)$ and $pq = \frac{c}{a} ...(2)$ .

Second equation, $a_1x² + b_1x + c_1 =0$ with $p_1$ and $(-q)$ as the roots.

Again, by Vieta's Formula, we have $p_1 - q = -\frac{b_1}{a_1} ...(3)$ and $-p_1q = \frac{c_1}{a_1} ...(4)$ .

$(1) + (3)$ gives us

$p + p_1 = -\frac{b}{a} - \frac{b_1}{a_1} = -\left( \frac{b}{a} + \frac{b_1}{a_1} \right) ...(5)$

Let's do something tricky. $(1) : (2) = \frac{1}{q} + \frac{1}{p} = -\frac{b}{a} \times \frac{a}{c}$

$\dfrac{1}{p} + \dfrac{1}{q} = -\frac{b}{c} ...(6)$ .

Another tricky thing. $(3) : (4) = -\dfrac{1}{q} + \frac{1}{p_1} = -\frac{b_1}{c_1}$

$-\frac{1}{q} + \frac{1}{p_1} = -\frac{b_1}{c_1} ...(7)$

$(6)+(7) \Rightarrow \frac{1}{p} + \frac{1}{p_1} = -\frac{b}{c} - \frac{b_1}{c_1} = -\left( \frac{b}{c} + \frac{b_1}{c_1} \right) ...(8)$ .

Now, note that $\frac{1}{p} + \frac{1}{p_1} = \frac{p + p_1}{pp_1}$ .

Then, from the equation $(8)$ , we have

$pp_1 = \left(\frac{\frac{b}{a} + \frac{b_1}{a_1}}{\frac{b}{c} + \frac{b_1}{c_1}} \right)$ .

Now, we have to find the roots of the equation $\frac{x²}{\frac{b}{a} + \frac{b_1}{a_1}} + x + \frac{1}{\frac{b}{c} + \frac{b_1}{c_1}} =0$ .

Multiply both sides with $\left(\frac{b}{a} + \frac{b_1}{a_1} \right)$ , we have

$x² + \left(\frac{b}{a} + \frac{b_1}{a_1} \right)x + \left( \frac{\frac{b}{a} + \frac{b_1}{a_1}}{\frac{b}{c} + \frac{b_1}{c_1}} \right) =0$

$x² - \left[ - \left( \frac{b}{a} + \frac{b_1}{a_1} \right)x \right] + \left( \frac{\frac{b}{a} + \frac{b_1}{a_1}}{\frac{b}{c} + \frac{b_1}{c_1}} \right) =0$

$x² - (p+p_1)x + (pp_1) = 0$

Now, we can clearly see that the roots are $p$ and $p_1$ .