Question

The condition that a roots of the equation $a{x}^2+bx+c=0$ may be reciprocal to the roots of $a_1{x}^2+b_{1}x+c_1=0$ is

• A. $(b{b}_1-a{a}_1{)}^2=(a{b}_1-b{c}_1\left)\right(b{a}_1-b_{1}c)$
• B. $(c{b}_1-b{a}_1{)}^2=(a{c}_1-c{d}_1\left)\right(a{b}_1-b{c}_1)$
• C. $(c{c}_1-a{a}_1{)}^2=(a{b}_1-b{c}_1\left)\right(b{a}_1-b_{1}c)$
• D. $a+b+c=0$

Answer 1

Answer: (C)

$(c{c}_1-a{a}_1{)}^2=(a{b}_1-b{c}_1\left)\right(b{a}_1-b_{1}c)$

Given $a{x}^2+bx+c=0$ ....(1)
and $a_1{x}^2+b_{1}x+c_1=0$ ....(2)
Replacing $x$ by $\frac{1}{x}$ in (2) we get
$c_1{x}^2+b_{1}x+a_1=0$ ....(3)
Let $\alpha$ be a common root of (1) and (3)
Solving (1) and (3), we get
$\alpha =\frac{c{c}_1-a{a}_1}{a{b}_1-b{c}_1}=\frac{b{a}_1-b_{1}c}{c{c}_1-a{a}_1}$
$\Rightarrow (c{c}_1-a{a}_1{)}^2=(a{b}_1-b{c}_1\left)\right(b{a}_1-b_{1}c)$
Hence, option 'C' is correct.