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Mathematics

If a2+1a2=11a^2 +\dfrac{1}{a^2}= 11, find: a−1aa -\dfrac{1}{a}

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Answer

Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

So,

(a−1a)2=a2−2×a×1a+(1a)2⇒(a−1a)2=a2−2+1a2\Big(a - \dfrac{1}{a}\Big)^2 = a^2 - 2 \times a \times \dfrac{1}{a} + \Big(\dfrac{1}{a}\Big)^2\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big)^2 = a^2 - 2 + \dfrac{1}{a^2}

Putting the value a2+1a2=11a^2 + \dfrac{1}{a}^2 = 11,we get

⇒(a−1a)2=(a2+1a2)−2⇒(a−1a)2=11−2⇒(a−1a)2=9⇒(a−1a)=9⇒(a−1a)=3 or−3⇒ \Big(a - \dfrac{1}{a}\Big)^2 = \Big(a^2 + \dfrac{1}{a^2}\Big) - 2\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big)^2 = 11 - 2\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big)^2 = 9\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big) = \sqrt{9}\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big) = 3 \text{ or} -3

Hence, the values of (a−1a)\Big(a - \dfrac{1}{a}\Big) are 3 or -3.

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