Question

If $a - \dfrac{1}{a} = 8$ and \[a \ne 0\]; find ${a^2} - \dfrac{1}{{{a^2}}}$
(A) $ \pm 11\sqrt {17} $
(B) $ \pm 13\sqrt {17} $
(C) $ \pm 16\sqrt {17} $
(D) $ \pm 17\sqrt {17} $

Answer 1

Hint:
Firstly find out the value of \[\left( {a + \dfrac{1}{a}} \right)\] by using $a - \dfrac{1}{a} = 8$ and then use formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ to find ${a^2} - \dfrac{1}{{{a^2}}}$ which is equal to $\left( {a + \dfrac{1}{a}} \right)\left( {a - \dfrac{1}{a}} \right)$.

Complete step by step solution:
Given, $a - \dfrac{1}{a} = 8$ ….. (1)
On squaring both sides of (1), we get-
${\left( {a - \dfrac{1}{a}} \right)^2} = {\left( 8 \right)^2}$
$ \Rightarrow $${a^2} + {\left( {\dfrac{1}{a}} \right)^2} - 2 \times a \times \dfrac{1}{a} = 64$ [Using ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$]
$ \Rightarrow $${a^2} + \dfrac{1}{{{a^2}}} - 2 = 64$
$ \Rightarrow $${a^2} + \dfrac{1}{{{a^2}}} = 64 + 2$
$ \Rightarrow $${a^2} + \dfrac{1}{{{a^2}}} = 66$ …. (2)
Now, ${\left( {a + \dfrac{1}{a}} \right)^2} = {a^2} + {\left( {\dfrac{1}{a}} \right)^2} + 2 \times a \times \dfrac{1}{a}$ [Using ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$]
$ \Rightarrow {\left( {a + \dfrac{1}{a}} \right)^2} = \left( {{a^2} + \dfrac{1}{{{a^2}}}} \right) + 2$
$ \Rightarrow {\left( {a + \dfrac{1}{a}} \right)^2} = 66 + 2$ [From (2)]
$ \Rightarrow {\left( {a + \dfrac{1}{a}} \right)^2} = 68$
On taking square root, we get-
$a + \dfrac{1}{a} = \sqrt {68} $
$ \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt {17} $ ….. (3)
We have to find the value of ${a^2} - \dfrac{1}{{{a^2}}}$.
${a^2} - \dfrac{1}{{{a^2}}} = \left( {a + \dfrac{1}{a}} \right)\left( {a - \dfrac{1}{a}} \right)$ [Using ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$]
$ \Rightarrow {a^2} - \dfrac{1}{{{a^2}}} = \pm 2\sqrt {17} \times 8$ [From (1)& (3)]
$ \Rightarrow {a^2} - \dfrac{1}{{{a^2}}} = \pm 16\sqrt {17} $

Hence, option (C) is the correct answer.

Note:
Note: An another method to solve this question is described below:
Given, $a - \dfrac{1}{a} = 8$ ….. (1)
We know that if $a - \dfrac{1}{a} = x$, then $a + \dfrac{1}{a} = \sqrt {{x^2} + 4} $
$\therefore a + \dfrac{1}{a} = \sqrt {{{\left( 8 \right)}^2} + 4} $
$ \Rightarrow a + \dfrac{1}{a} = \sqrt {64 + 4} $
$a + \dfrac{1}{a} = \sqrt {68} $
$ \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt {17} $ ….. (2)
We have to find the value of ${a^2} - \dfrac{1}{{{a^2}}}$.
 ${a^2} - \dfrac{1}{{{a^2}}} = \left( {a + \dfrac{1}{a}} \right)\left( {a - \dfrac{1}{a}} \right)$ [Using ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$]
$ \Rightarrow {a^2} - \dfrac{1}{{{a^2}}} = \pm 2\sqrt {17} \times 8$ [From (1)& (2)]
$ \Rightarrow {a^2} - \dfrac{1}{{{a^2}}} = \pm 16\sqrt {17} $
Hence, option (C) is the correct answer.