Question

If the sum of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is equal to the sum of the squares of their reciprocals, then prove that $2{{a}^{2}}c={{c}^{2}}b+{{b}^{2}}a$.

Answer 1

Hint: For solving this question, we let the roots of the polynomial equation $a{{x}^{2}}+bx+c=0$ be p and q. By using the condition given in the question, i.e. if the sum of roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is equal to the sum of the squares of their reciprocal, we get an equation in p and q. Now, by using the general form of sum and product of roots, we obtain the desired result.

Complete step-by-step solution -
The given quadratic equation is $a{{x}^{2}}+bx+c=0$. Let p and q are the roots of the given quadratic equations. As per given in the question, if the sum of roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is equal to the sum of the squares of their reciprocal. This can be mathematically expressed as:
$p+q=\dfrac{1}{{{p}^{2}}}+\dfrac{1}{{{q}^{2}}}...\left( 1 \right)$
As we know that the sum of the roots of quadratic equation is $\dfrac{-b}{a}$ where, b is coefficient of x and a is the coefficient of ${{x}^{2}},\left[ p+q=\dfrac{-b}{a} \right]$ and the product of the roots of quadratic equation is $\dfrac{c}{a}$ where, c is constant and a is the coefficient of ${{x}^{2}},\left[ pq=\dfrac{c}{a} \right]$.
From equation (1), the value of p + q is:
$\begin{aligned} & \Rightarrow p+q=\dfrac{1}{{{p}^{2}}}+\dfrac{1}{{{q}^{2}}} \\\ & \Rightarrow p+q=\dfrac{{{p}^{2}}+{{q}^{2}}}{{{p}^{2}}{{q}^{2}}} \\\ \end{aligned}$
Adding the subtracting 2pq in the numerator of right hand of the equation, we get:
$\Rightarrow p+q=\dfrac{{{p}^{2}}+{{q}^{2}}-2pq+2pq}{{{p}^{2}}{{q}^{2}}}$
We know that the ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$, using this identity we get:
$\Rightarrow p+q=\dfrac{{{\left( p+q \right)}^{2}}-2pq}{{{p}^{2}}{{q}^{2}}}$
Putting the value of $p+q=\dfrac{-b}{a}\text{ and }pq=\dfrac{c}{a}$ in the above equation, we get:
$\begin{aligned} & \Rightarrow \dfrac{-b}{a}=\dfrac{{{\left( \dfrac{-b}{a} \right)}^{2}}-\dfrac{2c}{a}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}} \\\ & \Rightarrow \dfrac{-b}{a}=\dfrac{\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{2c}{a}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}} \\\ & \Rightarrow \dfrac{-b}{a}=\dfrac{\dfrac{{{b}^{2}}-2ac}{{{a}^{2}}}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}} \\\ \end{aligned}$
Shifting the denominator of right-hand side in the left-hand side, we get:
$\begin{aligned} & \Rightarrow \dfrac{-b}{a}\times \dfrac{{{c}^{2}}}{{{a}^{2}}}=\dfrac{a{{b}^{2}}-2{{a}^{2}}c}{{{a}^{2}}\times a} \\\ & \Rightarrow \dfrac{-b{{c}^{2}}}{{{a}^{3}}}=\dfrac{a{{b}^{2}}-2{{a}^{2}}c}{{{a}^{3}}} \\\ \end{aligned}$
${{a}^{3}}$ is present in both sides of denominator so cancel out each other, we get
$\begin{aligned} & \Rightarrow -{{c}^{2}}b=a{{b}^{2}}-2{{a}^{2}}c \\\ & \Rightarrow {{c}^{2}}b+a{{b}^{2}}=2{{a}^{2}}c \\\ \end{aligned}$
Hence, we proved the equivalence of both sides.

Note: Students must first convert the problem statement into equation correctly and then the equation to desired form as a sum or product of roots to substitute the coefficients in the equation. Knowledge of sum and product of roots is must for solving this question.