Question

If each pair of equations $a_1{x}^2+b_{1}x+c_1=0,a_2{x}^2+b_{2}x+c_2=0$ and $a_3{x}^2+b_{3}x+c_3=0$ has a common root, then show that
(i) $\frac{c_1{a}_2+c_2{a}_1}{c_1{a}_2-c_2{a}_1}+\frac{a_1{a}_2{b}_3}{a_3(a_1{b}_2-a_2{b}_1)}=0$
(ii)${\left(\frac{a_1{b}_2-a_2{b}_1}{a_1{c}_2-a_2{c}_1}\right)}^2=\frac{a_1{a}_2{c}_3}{c_1{c}_2{a}_3}$

Answer 1

Since each pair has a common root, we choose them as

$\alpha ,\beta$ for I; $\beta ,\gamma$ for II and $\gamma ,\alpha$ for III

$\therefore$ $\alpha +\beta =-\frac{b_1}{a_1},\beta +\gamma =-\frac{b_2}{a_2},\gamma +\alpha =-\frac{b_3}{a_3}$ ........(A)

$\alpha \beta =\frac{c_1}{a_1},\beta \gamma =\frac{c_2}{a_2},\gamma \alpha =\frac{c_3}{a_3}$.........(B)

The first relation can be re-written as

(i) $\frac{a_1{a}_2(\frac{c_1}{a_1}+\frac{c_2}{a_2})}{a_1{a}_2(\frac{c_1}{a_1}-\frac{c_2}{a_2})}+\frac{b_3}{a_3}\frac{1}{\left(\frac{b_2}{a_2}\frac{b_1}{a_1}\right)}$

or $\frac{(\alpha \beta +\beta \gamma )}{(\alpha \beta -\beta \gamma )}-(\gamma +\alpha )\frac{1}{-(\beta +\gamma )+(\alpha +\beta )}$

or $\frac{\alpha +\gamma }{\alpha -\gamma }-\frac{\gamma +\alpha }{\alpha -\gamma }=0$

Again L.H.S. of 2nd relation is

(1) ${\left[\frac{a_1{a}_2\left(\frac{b_2}{a_2}\frac{b_1}{a_1}\right)}{a_1{a}_2(\frac{c_2}{a_1}-\frac{c_1}{a_1})}\right]}^2={\left[\frac{-(\beta +\gamma )+(\alpha +\beta )}{(\beta \gamma -\alpha \beta )}\right]}^2=\frac{{(\alpha -\gamma )}^2}{{\left[\beta \right(\gamma -\alpha \left)\right]}^2}=\frac{1}{{\beta }^2}$

R.H.S. $=\frac{c_3}{a_3}.\frac{1}{\frac{c_1}{a_1}.\frac{c_2}{a_2}}=\gamma \alpha .\frac{1}{\beta \gamma .\alpha \beta }=\frac{1}{{\beta }^2}$

Hence $L.H.S.=R.H.S.=\frac{1}{{\beta }^2}$