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Mathematics

If x+1xx + \dfrac{1}{x} = 4, find the values of

(i) x2+1x2x^2 + \dfrac{1}{x^2}

(ii) x4+1x4x^4 + \dfrac{1}{x^4}

(iii) x3+1x3x^3 + \dfrac{1}{x^3}

(iv) x−1xx - \dfrac{1}{x}.

Expansions

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Answer

(i) We know that,

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

Substituting values we get,

x2+1x2=42−2=16−2=14.x^2 + \dfrac{1}{x^2} = 4^2 - 2 \\[1em] = 16 - 2 \\[1em] = 14.

Hence, x2+1x2x^2 + \dfrac{1}{x^2} = 14.

(ii) We know that,

⇒(x+1x)2=x2+1x2+2⇒x2+1x2=(x+1x)2−2∴x4+1x4=(x2)2+1(x2)2=(x2+1x2)2−2.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2 \\[1em] \therefore x^4 + \dfrac{1}{x^4} = (x^2)^2 + \dfrac{1}{(x^2)^2} = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2.

Substituting values we get,

x4+1x4=142−2=196−2=194.x^4 + \dfrac{1}{x^4} = 14^2 - 2 \\[1em] = 196 - 2 \\[1em] = 194.

Hence, x4+1x4x^4 + \dfrac{1}{x^4} = 194.

(iii) We know that,

x3+1x3=(x+1x)3−3×x×1x(x+1x)x^3 + \dfrac{1}{x^3} = \Big(x + \dfrac{1}{x}\Big)^3 - 3 \times x \times \dfrac{1}{x}\Big(x + \dfrac{1}{x}\Big)

Substituting values we get,

x3+1x3=(4)3−3×4=64−12=52.x^3 + \dfrac{1}{x^3} = (4)^3 - 3 \times 4 \\[1em] = 64 - 12 \\[1em] = 52.

Hence, x3+1x3x^3 + \dfrac{1}{x^3} = 52.

(iv) We know that,

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

Substituting values we get,

x−1x=14−2=12=4×3=±23.x - \dfrac{1}{x} = \sqrt{14 - 2} \\[1em] = \sqrt{12} \\[1em] = \sqrt{4 \times 3} \\[1em] = \pm 2\sqrt{3}.

Hence, x−1x=±23x - \dfrac{1}{x} = \pm 2\sqrt{3}.

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