Question

If one root of $a{x}^2+bx+c=0$ is reciprocal of one root of $a_1{x}^2+b_{1}x+c_1=0$, then which of the following condition is correct?

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

Answer 1

The correct option is C $(a{a}_1-c{c}_1{)}^2=(a_{1}b-b_{1}c\left)\right(b_{1}a-b{c}_1)$

Let the equation
$a{x}^2+bx+c=0\cdots \left(1\right)$ has roots $\alpha ,\beta$
$a_1{x}^2+b_{1}x+c_1=0\cdots \left(2\right)$ has roots $\frac{1}{\alpha },\gamma$

Equation with roots $\frac{1}{\alpha },\frac{1}{\beta }$ is
$\frac{a}{x^2}+\frac{b}{x}+c=0\Rightarrow c{x}^2+bx+a=0\cdots \left(3\right)$

Equation $\left(2\right)$ and $\left(3\right)$ has one common root.
So, apply common root condition, we get
$(a{a}_1-c{c}_1{)}^2=(a_{1}b-b_{1}c\left)\right(b_{1}a-b{c}_1)$