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Mathematics

If x+1xx + \dfrac{1}{x} = 6, find

(i) x−1xx - \dfrac{1}{x}

(ii) x2−1x2x^2 - \dfrac{1}{x^2}

Expansions

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Answer

(i) We know that,

⇒(x−1x)2=x2+1x2−2 ……(i)⇒(x+1x)2=x2+1x2+2 ……(ii)\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \text{ ……(i)} \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \text{ ……(ii)}

Eq. (i) can be written as,

⇒(x−1x)2=x2+1x2+2−4⇒(x−1x)2=(x+1x)2−4∴x−1x=(x+1x)2−4\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 - 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \Big(x + \dfrac{1}{x}\Big)^2 - 4 \\[1em] \therefore x - \dfrac{1}{x} = \sqrt{\Big(x + \dfrac{1}{x}\Big)^2 - 4}

Substituting values we get,

x−1x=(6)2−4=36−4=32=±42.x - \dfrac{1}{x} = \sqrt{(6)^2 - 4} \\[1em] = \sqrt{36 - 4} \\[1em] = \sqrt{32} \\[1em] = \pm 4\sqrt{2}.

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

(ii) We know that,

x2−1x2=(x+1x)(x−1x).x^2 - \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big).

When, x−1x=42x - \dfrac{1}{x} = 4\sqrt{2}

Substituting values we get,

x2−1x2=6×42=242.x^2 - \dfrac{1}{x^2} = 6 \times 4\sqrt{2} = 24\sqrt{2}.

When, x−1x=−42x - \dfrac{1}{x} = -4\sqrt{2}

Substituting values we get,

x2−1x2=6×−42=−242.x^2 - \dfrac{1}{x^2} = 6 \times -4\sqrt{2} = -24\sqrt{2}.

Hence, x2−1x2=±242.x^2 - \dfrac{1}{x^2} = ±24\sqrt{2}.

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