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

From the given quadratic equations, we have

${\alpha }_1{\beta }_1=\frac{c_1}{a_1},{\alpha }_2{\beta }_2=\frac{c_2}{a_2}$

${\alpha }_1+{\beta }_1=\frac{-b_1}{a_1},{\alpha }_2+{\beta }_2=\frac{-b_2}{a_2}$

${\alpha }_1{\beta }_1{\alpha }_2{\beta }_2=\frac{c_1}{a_1}\times \frac{c_2}{a_2}=1$

So, $\frac{c_1}{a_1}=\frac{a_2}{c_2}$

${\alpha }_1{\beta }_{1\times }({\alpha }_2+{\beta }_2)=\frac{c_1}{a_1}\times \frac{-b_2}{a_2}$

$\frac{c_1}{a_2}=\frac{b_1}{b_2}$

Hence, option 'B' is correct.