Question

Find the value of ‘k’ for which the roots of the equation $3{{x}^{2}}-10x+k=0$ are reciprocal of each other.

Answer 1

Hint: Assume that one root the equation is ‘u’ and thus, the other root is $\dfrac{1}{u}$. Use the fact that the sum and product of roots of the equation of the form $a{{x}^{2}}+bx+c=0$ are $\dfrac{-b}{a}$ and $\dfrac{c}{a}$ respectively. Write equations based on the data given in the question and simplify them to find the value of variable ‘k’.

Complete step-by-step solution -
We have to calculate the value of variable ‘k’ such that the roots of the equation $3{{x}^{2}}-10x+k=0$ are reciprocal of each other.
Let’s assume that one root of the given quadratic equation is ‘u’. As both the roots are reciprocal of each other, the other root of the quadratic equation is $\dfrac{1}{u}$.
We know that the sum and product of roots of the equation of the form $a{{x}^{2}}+bx+c=0$ are $\dfrac{-b}{a}$ and $\dfrac{c}{a}$ respectively.
Substituting $a=3,b=-10,c=k$ in the above expression, the sum and product of the roots of the equation $3{{x}^{2}}-10x+k=0$ are $\dfrac{10}{3}$ and $\dfrac{k}{3}$ respectively.
We know that the roots of the equation $3{{x}^{2}}-10x+k=0$ are u and $\dfrac{1}{u}$.
Thus, we have $u+\dfrac{1}{u}=\dfrac{10}{3}.....\left( 1 \right)$ and $u\left( \dfrac{1}{u} \right)=\dfrac{k}{3}.....\left( 2 \right)$.
Simplifying equation (2), we have $u\left( \dfrac{1}{u} \right)=1=\dfrac{k}{3}$.
Cross multiplying the terms of the above equation, we have $k=1\times 3=3$.
Hence, the value of ‘k’ for which the roots of the equation $3{{x}^{2}}-10x+k=0$ are reciprocal of each other is $k=3$.

Note: We can calculate the exact roots of the equation $3{{x}^{2}}-10x+k=0$ by simplifying the equation $u+\dfrac{1}{u}=\dfrac{10}{3}$. We can also calculate the exact roots of the given quadratic equation by substituting the value of ‘k’ in the equation and factorising it by splitting the middle term method.