Question

$\alpha$ and $\beta$ are the roots of $a{x}^2+bx+c=0$ and $a{x}^2+bx+c=0$, then prove that $f\left(\alpha \right).f\left(\beta \right)=\frac{1}{a^2}[{(c_{1}a-c{a}_1)}^2-(a{b}_1-a_{1}b)(b{c}_1-b_{1}c)]$

Answer 1

$f\left(\alpha \right)=a_1{\alpha }^2+b_1\alpha +c_1$
or $a_1{\alpha }^2+b_1\alpha +[c_1-f\left(\alpha \right)]=0$.......(1)
Again $\alpha$ is a root of $a{x}^2+bx+c=0$
$\therefore$ $a{\alpha }^2+bx+c=0$........(2)
solving (1) and (2) by cross multiplication, we get
$\frac{{\alpha }^2}{b_{1}c-b{c}_1+bf\left(\alpha \right)}=\frac{\alpha }{a{c}_1-c{a}_1-af\left(\alpha \right)}=\frac{1}{a_{1}b-a{b}_1}$
$\therefore$ ${[a{c}_1-a_{1}c-af(\alpha \left)\right]}^2=(a_{1}b-a{b}_1)[b_{1}c-b{c}_1+bf(\alpha \left)\right]$
Writing above as a quadratic in $f\left(\alpha \right)$
$a^2{\left[f\right(\alpha \left)\right]}^2+f\left(\alpha \right)\left[\right]+{(a{c}_1-a_{1}c)}^2-(a_{1}b-a{b}_1)(b_{1}c-b{c}_1)=0$
Above shows that $f\left(\alpha \right)$ is a root of
$a^2{t}^2+t\left[\right]+{(a{c}_1-a_{1}c)}^2-(a_{1}b-a{b}_1)(b_{1}c-b{c}_1)=0$...........(1)
Similarly was can assume that $f\left(\beta \right)$ will also be a root of (1). Hence the roots of t quadratic given by (1) are $f\left(\alpha \right)$ and $f\left(\beta \right)$.
$\therefore$ $f\left(\alpha \right).f\left(\beta \right)$= Product of roots of (1).
or $f\left(\alpha \right).f\left(\beta \right)=\frac{1}{a^2}\left[{(a{c}_1-a_{1}c)}^2(a_{1}b-a{b}_1)(b_{1}c-b{c}_1)\right]$........(2)
It is same as given result if we change the signs of both the last brackets in the numerator.
In case $\alpha$ or $\beta$ be a common root of
$a{x}^2+bx+c=0$ and $f\left(x\right)=a_1{x}^2+b^{}1x+c=0$
then $f\left(\alpha \right)=0$.
Hence from [2], $
f(\alpha ).f(\beta )=0$
$\therefore$ ${(a{c}_1-a_{1}c)}^2-(a_{1}b-a{b}_1)(b_{1}c-b{c}_1)=0$
or ${(a{c}_1-a_{1}c)}^2=(b{c}_1-b_{1}c)(a{b}_1-a_{1}b)$
is the required condition.