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# Find the value of $a - \dfrac{1}{a}$ in terms of ‘$m$’, if $a + \dfrac{1}{a} = m$and$a \ne 0$.A). $\pm \sqrt {{m^2} - 4}$B). $\pm \sqrt {{m^2} - 3}$C). $\pm \sqrt {{m^2} - 9}$D). $\pm \sqrt {{m^2} - 1}$

Last updated date: 25th Mar 2023
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Hint: We need to use a given expression to find the value of another expression. We have to simplify a given expression and then find the value of the required expression in terms of ‘$m$’. We need to perform appropriate operations in order to have common terms,
Formula used:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$

Complete step-by-step solution:
Let us note the given equation,
$a + \dfrac{1}{a} = m$
Let us simplify the above equation,
On squaring both sides we get,
${\left( {a + \dfrac{1}{a}} \right)^2} = {m^2}$
On expanding the bracket using formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ we get,
${a^2} + 2 + \dfrac{1}{{{a^2}}} = {m^2}$
On subtraction $2$ from both sides we get,
${a^2} + \dfrac{1}{{{a^2}}} = {m^2} - 2$ $...................[1]$
We need to find the value of $a - \dfrac{1}{a}$
Let us simplify the above equation,
$a - \dfrac{1}{a}$
On squaring both sides we get,
${\left( {a - \dfrac{1}{a}} \right)^2} = {a^2} - 2 + \dfrac{1}{{{a^2}}}$
On rearranging the terms on R.H.S. we get,
${\left( {a - \dfrac{1}{a}} \right)^2} = {a^2} + \dfrac{1}{{{a^2}}} - 2$
Let us put the value from equation $[1]$ , ${a^2} + \dfrac{1}{{{a^2}}} = {m^2} - 2$ in above equation,
${\left( {a - \dfrac{1}{a}} \right)^2} = {m^2} - 2 - 2$
On performing subtraction on R.H.S. we get,
${\left( {a - \dfrac{1}{a}} \right)^2} = {m^2} - 4$
On taking square roots on the both sides we get,
$\therefore a - \dfrac{1}{a} = \pm \sqrt {{m^2} - 4}$
This is the required value.
Hence option A) $\pm \sqrt {{m^2} - 4}$is correct.

Note: In such questions where both terms have opposite signs, we need to perform operations like squaring, in order to get common terms from two different terms. Then we can substitute the value for a common term to obtain the required equation.