Question

If a root of the equation $a x^{2}+b x+c=0$ be reciprocal of a root of the equation $a^{\prime} x^{2}+b^{\prime} x+c^{\prime}=0$, then?
A. $\left(c c^{\prime}-a a^{\prime}\right)^{2}=\left(b a^{\prime}-c b^{\prime}\right)\left(a b^{\prime}-b c^{\prime}\right)$
B. $\left(b b^{\prime}-a a^{\prime}\right)^{2}=\left(c a^{\prime}-b c^{\prime}\right)\left(a b^{\prime}-b c^{\prime}\right)$
C. $\left(c c^{\prime}-a a^{\prime}\right)^{2}=\left(b a^{\prime}+c b^{\prime}\right)\left(a b^{\prime}+b c^{\prime}\right)$
D. None of the above

Answer 1

Hint: In this question, we have to find the relationship between the coefficients of the given two quadratic equations. We are given the relationship between the roots of these two quadratic equations. Here we are to use the substitution and elimination methods to reach the solution.

Complete step by step solution: We are given two quadratic equations — $a x^{2}+b x+c=0$ and $a^{\prime} x^{2}+b^{\prime} x+c^{\prime}=0$.
Let $\alpha$ and $\beta$ be the roots of the equation $a x^{2}+b x+c=0$ and $\gamma$ and $\delta$ be the roots of the equation $a^{\prime} x^{2}+b^{\prime} x+c^{\prime}=0$.
We have that one of the roots of $a^{\prime} x^{2}+b^{\prime} x+c^{\prime}=0$ is the reciprocal of a root of the equation $a x^{2}+b x+c=0$.
Therefore, we have
$\delta=\dfrac{1}{\alpha}$
Therefore, the roots of the equation $a^{\prime} x^{2}+b^{\prime} x+c^{\prime}=0$ are $\gamma$ and $\dfrac{1}{\alpha}$.
Now we have $\alpha$ is a root of the equation $a x^{2}+b x+c=0$, then
$a \alpha^{2}+b \alpha+c=0$ ----- (1)
Also, we have $\dfrac{1}{\alpha}$ is a root of the equation $a^{\prime} x^{2}+b^{\prime} x+c^{\prime}=0$, then
$a^{\prime}\left(\dfrac{1}{\alpha}\right)^{2}+b^{\prime}\left(\dfrac{1}{\alpha}\right)+c^{\prime}=0$
$\dfrac{a^{\prime}}{\alpha^{2}}+\dfrac{b^{\prime}}{\alpha}+c^{\prime}=0 $
$\dfrac{a^{\prime}+b^{\prime} \alpha+c^{\prime} \alpha^{2}}{\alpha^{2}}=0$
$c^{\prime} \alpha^{2}+b^{\prime} \alpha+a^{\prime}=0$ ------ (2)
Multiply equation (1) by $c^{\prime}$ and equation (2) by $a$.
$c^{\prime} a \alpha^{2}+c^{\prime} b \alpha+c^{\prime} c=0$
$c^{\prime} a \alpha^{2}+b^{\prime} a \alpha+a^{\prime} a=0$
Subtracting these two equations we get,
$c^{\prime} a \alpha^{2}+c^{\prime} b \alpha+c^{\prime} c=0$
$c^{\prime} a \alpha^{2}+b^{\prime} a \alpha+a^{\prime} a=0$

$\left(c^{\prime} b-b^{\prime} a\right) \alpha+\left(c^{\prime} c-a^{\prime} a\right)=0$
$\alpha=\dfrac{a a^{\prime}-c c^{\prime}}{c^{\prime} b-b^{\prime} a}$
Substitute the value of in equation (1),
$a\left[\dfrac{a a^{\prime}-c c^{\prime}}{c^{\prime} b-b^{\prime} a}\right]^{2}+b\left[\dfrac{a a^{\prime}-c c^{\prime}}{c^{\prime} b-b^{\prime} a}\right]+c=0 $
$\dfrac{a\left(a a^{\prime}-c c^{\prime}\right)^{2}+b\left(a a^{\prime}-c c^{\prime}\right)\left(c^{\prime} b-b^{\prime} a\right)+c\left(c^{\prime} b-b^{\prime} a\right)^{2}}{\left(c^{\prime} b-b^{\prime} a\right)^{2}}=0$
$a\left(a a^{\prime}-c c^{\prime}\right)^{2}+b\left(a a^{\prime}-c c^{\prime}\right)\left(c^{\prime} b-b^{\prime} a\right)+c\left(c^{\prime} b-b^{\prime} a\right)^{2}=0$
$a\left(a a^{\prime}-c c^{\prime}\right)^{2}+\left(c^{\prime} b-b^{\prime} a\right)\left[b\left(a a^{\prime}-c c^{\prime}\right)+c\left(c^{\prime} b-b^{\prime} a\right)\right]=0$
$a\left(a a^{\prime}-c c^{\prime}\right)^{2}+\left(c^{\prime} b-b^{\prime} a\right)\left[b a a^{\prime}-b c c^{\prime}+c c^{\prime} b-b^{\prime} a c\right]=0$
$a\left(a a^{\prime}-c c^{\prime}\right)^{2}+\left(c^{\prime} b-b^{\prime} a\right)\left(b a^{\prime}-b^{\prime} a c\right)=0$
$a\left(a a^{\prime}-c c^{\prime}\right)+a\left(c^{\prime} b-b^{\prime} a\right)\left(b a^{\prime}-b^{\prime} c\right)=0$
$a\left[\left(a a^{\prime}-c c^{\prime}\right)^{2}+\left(c^{\prime} b-b^{\prime} a\right)\left(a^{\prime} b-b^{\prime} c\right)\right]=0$
$\left(a a^{\prime}-c c^{\prime}\right)^{2}+\left(c^{\prime} b-b^{\prime} a\right)\left(a ^{\prime} b-b^{\prime} c\right)=0$
$\left(a a^{\prime}-c c^{\prime}\right)^{2}=-\left[\left(c^{\prime} b-a b^{\prime}\right)\left(a^{\prime} b-c b^{\prime}\right)\right]$
$\left(c c^{\prime}-a a^{\prime}\right)^{2}=\left(c^{\prime} b-a b^{\prime}\right)\left(a ^{\prime} b-c b^{\prime}\right)$

So, Option ‘A’ is correct

Note: In this question even though we were dealing with the roots of the polynomial we did not use the relationship between the roots and the coefficients. We have used the basic concepts of roots and substituted the values we obtained. Be careful with the substitution part, do not expand all the brackets instead look for common terms, bring them together then simply the terms.