Question

If the roots of $a_1{x}^2+b_{1}x+c_1=0$ are ${\alpha }_1,{\beta }_1,$ and those of $a_2{x}^2+b_{2}x+c_2=0$ are ${\alpha }_2,{\beta }_2$ such that ${\alpha }_1{\alpha }_2={\beta }_1{\beta }_2=1$, then

• A. $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
• B. $\frac{a_1}{c_2}=\frac{b_1}{b_2}=\frac{c_1}{a_2}$
• C. $a_1{a}_2=b_1{b}_2=c_1{c}_2$
• D. None of these

Answer 1

The correct option is A $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$a_1{x}^2+b_{1}x+c_1=0$

${\alpha }_1+{\beta }_1=\frac{-b_1}{a_1}$

${\alpha }_1{\beta }_1=\frac{c_1}{a_1}$

$a_2{x}^2+b_{2}x+c_2=0$

${\alpha }_2+{\beta }_2=\frac{-b_2}{a_2}$

${\alpha }_2{\beta }_2=\frac{c_2}{a_2}$

${\alpha }_1{\alpha }_2={\beta }_1{\beta }_2=1$

$\therefore {\alpha }_1{\beta }_1{\alpha }_2{\beta }_2=\left({\alpha }_1{\alpha }_1\right)\cdot \left({\beta }_1{\beta }_1\right)=1\cdot 1=1$

$\therefore {\alpha }_1{\beta }_1{\alpha }_2{\beta }_2=\left({\alpha }_1{\alpha }_1\right)\cdot \left({\beta }_1{\beta }_1\right)=\frac{c_1}{a_1}\cdot \frac{c_2}{a_2}=1$

$⟹\frac{c_1}{a_2}=\frac{a_1}{c_2}$

Now, ${\alpha }_1+{\beta }_1=\frac{-b_1}{a_1}$

$⟹\frac{1}{{\alpha }_2}+\frac{1}{{\beta }_2}=\frac{-b_1}{a_1}$

$⟹\frac{{\alpha }_2+{\beta }_2}{{\alpha }_2{\beta }_2}=\frac{-b_1}{a_1}$

$⟹\frac{-b_2/a_2}{c_2/a_2}=\frac{-b_1}{a_1}$

$⟹\frac{b_2}{c_2}=\frac{b_1}{a_1}$

$⟹\frac{a_1}{c_2}=\frac{b_1}{b_2}$

$\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$