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Mathematics

If a+1a=6a + \dfrac{1}{a} = 6 and a ≠ 0; find :

(i) a−1aa - \dfrac{1}{a}

(ii) a2−1a2a^2 - \dfrac{1}{a^2}

Expansions

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Answer

(i) By formula,

⇒(a−1a)2=a2+1a2−(2×a×1a)⇒(a−1a)2=a2+1a2−2⇒(a−1a)2=a2+1a2+2−4⇒(a−1a)2=(a+1a)2−4⇒(a−1a)2=62−4⇒(a−1a)2=36−4⇒(a−1a)2=32⇒a−1a=32=±42.\Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - \Big(2 \times a \times \dfrac{1}{a}\Big) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 - 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = \Big(a + \dfrac{1}{a}\Big)^2 - 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 6^2 - 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 36 - 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 32 \\[1em] \Rightarrow a - \dfrac{1}{a} = \sqrt{32} = \pm 4\sqrt{2}.

Hence, a−1a=±42a - \dfrac{1}{a} = \pm 4\sqrt{2}.

(ii) Solving,

⇒a2−1a2⇒(a−1a)(a+1a)⇒±42×6⇒±242.\Rightarrow a^2 - \dfrac{1}{a^2} \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)\Big(a + \dfrac{1}{a}\Big) \\[1em] \Rightarrow \pm 4\sqrt{2} \times 6 \\[1em] \Rightarrow \pm 24\sqrt{2}.

Hence, a2−1a2=±242a^2 - \dfrac{1}{a^2} = \pm 24\sqrt{2}.

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