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Mathematics

If x−2=13xx - 2 = \dfrac{1}{3x}, find the values of :

(i) (x2+19x2)\Big(x^2 + \dfrac{1}{9x^2}\Big)

(ii) (x4+181x4)\Big(x^4 + \dfrac{1}{81x^4}\Big).

Expansions

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Answer

(i) Given,

⇒x−2=13x⇒x−13x=2\Rightarrow x - 2 = \dfrac{1}{3x} \\[1em] \Rightarrow x - \dfrac{1}{3x} = 2

We know that,

⇒(x−13x)2=x2+(13x)2−2×x×13x⇒(2)2=x2+(13x)2−2×x×13x⇒4=x2+19x2−23⇒x2+19x2=4+23⇒x2+19x2=12+23⇒x2+19x2=143\Rightarrow \Big(x - \dfrac{1}{3x}\Big)^2 = x^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times x \times \dfrac{1}{3x} \\[1em] \Rightarrow (2)^2 = x^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times x \times \dfrac{1}{3x} \\[1em] \Rightarrow 4 = x^2 + \dfrac{1}{9x^2} - \dfrac{2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = 4 + \dfrac{2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{12 + 2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}

Hence, x2+19x2=143x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}.

(ii) From part (i),

x2+19x2=143x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}

Using identity,

⇒(x2+19x2)2=(x2)2+(19x2)2+2×x2×19x2⇒(143)2=(x2)2+(19x2)2+2×x2×19x2⇒1969=x4+181x4+29⇒x4+181x4=1969−29⇒x4+181x4=196−29⇒x4+181x4=1949\Rightarrow \Big(x^2 + \dfrac{1}{9x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{9x^2} \\[1em] \Rightarrow \Big(\dfrac{14}{3}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{9x^2} \\[1em] \Rightarrow \dfrac{196}{9} = x^4 + \dfrac{1}{81x^4} + \dfrac{2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{196}{9} - \dfrac{2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{196 - 2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{194}{9}

Hence, x4+181x4=1949x^4 + \dfrac{1}{81x^4} = \dfrac{194}{9}.

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