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If one root of $5{{x}^{2}}-13x+k=0$is reciprocal of the other, then
(a) k=0
(b) k=5
(c) $k=\dfrac{1}{6}$
(d) $k=6$

Answer Verified Verified
Hint: We will need to remember the relation of the roots of the equation with the coefficients used in the equation.
The roots $\alpha $ and \[\beta \] of the quadratic equation $a{{x}^{2}}+bx+c=0$ …(i) are related to the coefficients a, b and c in following manner
$\begin{align}
  & \alpha +\beta =-\dfrac{b}{a} \\
 & \alpha \beta =\dfrac{c}{a} \\
\end{align}$…(ii)
Use these equations to get the value of k.
The roots $\alpha $ and \[\beta \] of the quadratic equation $a{{x}^{2}}+bx+c=0$ …(i) are related to the coefficients a, b and c in following manner
$\begin{align}
  & \alpha +\beta =-\dfrac{b}{a} \\
 & \alpha \beta =\dfrac{c}{a} \\
\end{align}$…(ii)
Let us assume that our roots are also $\alpha $ and $\beta $ .
Since, one root is reciprocal of other we can write, $\beta =\dfrac{1}{\alpha }$
Therefore, we can write $\alpha \beta =1$ .
From equation (ii) we can write, $\alpha \beta =\dfrac{c}{a}=1$ . Now comparing the given equation with equation (i) we determine the value of a, b and c as
$\begin{align}
  & a=5 \\
 & b=13 \\
 & c=k \\
\end{align}$
Substituting these values in equation (ii) we have,
 $\begin{align}
  & \dfrac{k}{5}=1 \\
 & \Rightarrow k=5 \\
\end{align}$
Hence, the option (b) is the correct answer.

Note: This is the easiest and fastest way to solve this problem. Any other way will be unnecessarily long. In this problem one might overthink and might try to use the relation $\alpha +\beta =-\dfrac{b}{a}$ which is unnecessary and futile. But there are problems that require the value of both α and $\beta $. Then we will require a minimum of two conditions since we have two variables. In this question we were able to obtain the answer without using the above relation but the requirement will vary from question to question. But one thing we must try to do is solve the question by the easiest way and for that prior inspection will be needed rather than diving into equations because it will become cumbersome if we encounter lengthy questions.

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