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Mathematics

If (x+1x)=6\Big(x + \dfrac{1}{x}\Big) = 6, find the values of :

(i) (x−1x)\Big(x - \dfrac{1}{x}\Big).

(ii) (x2−1x2)\Big(x^2 - \dfrac{1}{x^2}\Big)

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Answer

(i) Given,

(x+1x)=6\Big(x + \dfrac{1}{x}\Big) = 6

We know that,

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

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

(ii) Given,

(x+1x)=6\Big(x + \dfrac{1}{x}\Big) = 6

From part (i),

⇒(x−1x)=±42\Rightarrow \Big(x - \dfrac{1}{x}\Big) = ±4\sqrt{2}

We know that,

⇒(x2−1x2)=(x+1x)(x−1x)⇒(x2−1x2)=6×±42⇒(x2−1x2)=±242\Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = 6 \times \pm 4\sqrt{2} \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \pm 24\sqrt{2}

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

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