Question

If one root of the equation \[\left( {{k^2} + 1} \right){x^2} + 13x + 4k = 0\] is reciprocal of the other, then k has the value:
A) $ - 2 + \sqrt 3 $
B) $2 - \sqrt 3 $
C) 1
D) None of these

Answer 1

Hint: We will first assume one of its roots and then find another one according to the given condition and now since we know the product of roots in a general equation, we will get the answer.

Complete step-by-step solution:
Let us assume one of the roots of the equation \[\left( {{k^2} + 1} \right){x^2} + 13x + 4k = 0\] is $a$.
Since we are given that the other root is reciprocal of this one. Therefore, the other root will be $\dfrac{1}{a}$.
Now, we have both the roots with us.
The general quadratic equation is given by $a{x^2} + bx + c = 0$.
Here, the sum of both the roots is given by $ - \dfrac{b}{a}$ and the product of the roots is given by $\dfrac{c}{a}$.
Therefore, the product of roots of \[\left( {{k^2} + 1} \right){x^2} + 13x + 4k = 0\] will be $\dfrac{{4k}}{{{k^2} + 1}}$. ………….(1)
Now, we can also see that the roots are $a$ and $\dfrac{1}{a}$. Their product = $a \times \dfrac{1}{a} = 1$. …………….(2)
Using the equations (1) and (2), we will get:-
$ \Rightarrow \dfrac{{4k}}{{{k^2} + 1}} = 1$
Cross – multiplying both the sides to get the following expression:-
$ \Rightarrow 4k = {k^2} + 1$
Re – arranging the terms to get the following expression:-
$ \Rightarrow {k^2} - 4k + 1 = 0$ ………………(3)
We know that the solutions of the general quadratic equation given by $a{x^2} + bx + c = 0$ are:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing the general equation to the equation (3), we have with us: a = 1, b = -4, c = 1.
$\therefore x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}$
Simplifying it to get the following expression:-
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 - 4} }}{2}$
Simplifying it further to get the following expression:-
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {12} }}{2}$
We can rewrite this expression as follows:-
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {2 \times 2 \times 3} }}{2}$
Taking 2 out of the square root, we will get:-
$ \Rightarrow x = \dfrac{{4 \pm 2\sqrt 3 }}{2}$
Rewriting the above expression as follows:-
$ \Rightarrow x = \dfrac{{2\left( {2 \pm \sqrt 3 } \right)}}{2}$
Simplifying the above expression to finally get the roots as following:-
$ \Rightarrow x = 2 \pm \sqrt 3 $

$\therefore $ The correct option is (B).

Note: The students must note that they may try to put in both the roots individually in the given quadratic equation and then solve them to get the required answer but that will create a lot of chaos. The way done above was the easiest as in comparison to all.
The students may check their answer by checking that these roots are the reciprocal of each other:-
$ \Rightarrow \dfrac{1}{{2 - \sqrt 3 }} = \dfrac{1}{{2 - \sqrt 3 }} \times \dfrac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }}$
\[ \Rightarrow \dfrac{1}{{2 - \sqrt 3 }} = \dfrac{{2 + \sqrt 3 }}{{{2^2} - {{\left( {\sqrt 3 } \right)}^2}}}\] (Because ${a^2} - {b^2} = (a + b)(a - b)$)
\[ \Rightarrow \dfrac{1}{{2 - \sqrt 3 }} = \dfrac{{2 + \sqrt 3 }}{{4 - 3}} = 2 + \sqrt 3 \]