$r$ and $s$ are roots of $ax^2+bx+c=0$ so what is the value of $\frac{1}{r^2}+\frac{1}{s^2}$ in terms of a b and c

Question

Problem The problems asks us in the equation $ax^2+bx+c=0$ and we need to compute $\frac{1}{r^2}+\frac{1}{s^2}$ in terms of a,b, and or c

My steps In this problem we have to use the factor theorem which states that: $P(x)=Q(x)*D(x)+R(x)$ where P(x) is the equation: $ax^2+bx+c=0$ and d(x) is the divisor and r(x) is the remainder and q(x) is the quotient

So my steps was to first set the two roots of P(x) as in terms of r and s

Though my steps was super short, i really don't know how to go from here and I got stuck

Answer 1

$ax^2+bx+c = a(x-r)(x-s)$. Comparing coefficients, we get $r+s = -\frac{b}{a}$ and $rs = \frac{c}{a}$. \begin{align*} \frac{1}{r^2} + \frac{1}{s^2} &= \frac{r^2+s^2}{r^2s^2}\\ &= \frac{(r+s)^2 - 2rs}{r^2s^2}\\ &= \frac{\frac{b^2}{a^2} - 2\frac{c}{a}}{\frac{c^2}{a^2}}\\ &=\frac{b^2-2ac}{c^2} \end{align*}

Answer 2

$$\frac 1 {r^2}+\frac 1 {s^2}=\frac {r^2+s^2} {r^2s^2}$$ We know that $$r^2=\frac {-br-c} a$$ and similar result for $s$.

Hence, we get $$\frac {r^2+s^2} {r^2s^2}=\frac {-b(r+s)-2c} {ar^2s^2}$$ $$=\frac {b^2-2ac} {a^2r^2s^2}$$ $$=\frac {b^2-2ac} {c^2}$$

since $$r+s=-\frac b a$$ and $$rs=\frac c a$$

Answer 3

You can also deal with the algebra involved in this manner:

The roots of the quadratic equation can be written as $ \ r \ = \ \alpha \ + \ \beta \ $ and $ \ s \ = \ \alpha \ - \ \beta \ $ , with $ \ \alpha \ = \ -\frac{b}{2a} \ $ and $ \ \beta \ = \ \frac{\sqrt{D}}{2a} \ $ , $ \ D \ $ being the discriminant $ \ b^2 \ - \ 4 a c \ $ . So $ \ \alpha^2 \ = \ \frac{b^2}{4 \ a^2} \ $ and $ \ \beta^2 \ = \ \frac{D }{4 \ a^2} \ = \ \frac{b^2 \ - \ 4 \ a \ c}{4 \ a^2} \ $ .

Thus, we have

$$ \ \frac{1}{r^2} \ + \ \frac{1}{s^2} \ \ = \ \ \left(\frac{1}{\alpha \ + \ \beta} \right)^2 \ + \ \left(\frac{1}{\alpha \ - \ \beta} \right)^2 \ \ \ = \ \ \frac{(\alpha \ - \ \beta)^2 \ + \ (\alpha \ + \ \beta)^2}{ ( \alpha \ + \ \beta )^2 \ ( \alpha \ - \ \beta )^2 } \ \ = \ \ \frac{2 \ \alpha^2 \ + \ 2 \ \beta^2}{ ( \alpha^2 \ - \ \beta^2 )^2 }$$

[applying the binomial theorem and the "difference of two squares"]

$$ = \ \ 2 \ \cdot \ \frac{ \left(\frac{b^2}{4 \ a^2} \right) \ + \ \ \left(\frac{b^2 \ - \ 4 \ a \ c}{4 \ a^2} \right)}{ \left[ \ \left(\frac{b^2}{4 \ a^2} \right) \ - \ \left(\frac{b^2 \ - \ 4 \ a \ c}{4 \ a^2} \right) \ \right]^2 } \ \ = \ \ 2 \ \cdot \ \frac{ \left(\frac{2b^2 \ - \ 4 \ a \ c}{4 \ a^2} \right)}{ \left( \frac{ 4 \ a \ c}{4 \ a^2} \right) ^2 } $$

$$ = \ \ 2 \ \cdot \ \left(\frac{2b^2 \ - \ 4 \ a \ c}{4 \ a^2} \right) \ \cdot \ \left( \frac{a}{c} \right)^2 \ \ = \ \ \frac{2b^2 \ - \ 4 \ a \ c}{2 \ c^2} \ \ = \ \ \frac{ b^2 \ - \ 2 \ a \ c}{ c^2} \ \ . $$