KnowledgeBoat Logo
|

Mathematics

If 2(x2 + 1) = 5x, find :

(i) x−1xx - \dfrac{1}{x}

(ii) x3−1x3x^3 - \dfrac{1}{x^3}

Expansions

29 Likes

Answer

(i) Given,

⇒2(x2+1)=5x⇒x2+1x=52⇒x+1x=52.\Rightarrow 2(x^2 + 1) = 5x \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{5}{2} \\[1em] \Rightarrow x + \dfrac{1}{x} = \dfrac{5}{2}.

By formula,

⇒(x+1x)2−(x−1x)2=4⇒(52)2−(x−1x)2=4⇒(x−1x)2=254−4⇒(x−1x)2=25−164⇒(x−1x)2=94⇒(x−1x)=94⇒(x−1x)=±32.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(\dfrac{5}{2}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{25}{4} - 4\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{25 - 16}{4} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{9}{4} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm \dfrac{3}{2}.

Hence, (x−1x)=±32.\Big(x - \dfrac{1}{x}\Big) = \pm \dfrac{3}{2}.

(ii) By formula,

⇒(x3−1x3)=(x−1x)3+3(x−1x)\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big) \\[1em]

Substituting (x−1x)=32\Big(x - \dfrac{1}{x}\Big) = \dfrac{3}{2}

⇒(x3−1x3)=(32)3+3×32=278+92=27+368=638.\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(\dfrac{3}{2}\Big)^3 + 3 \times \dfrac{3}{2} \\[1em] = \dfrac{27}{8} + \dfrac{9}{2} \\[1em] = \dfrac{27 + 36}{8} \\[1em] = \dfrac{63}{8}.

Substituting (x−1x)=−32\Big(x - \dfrac{1}{x}\Big) = -\dfrac{3}{2}

⇒(x3−1x3)=(−32)3+3×−32=−278−92=−27−368=−638.\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(-\dfrac{3}{2}\Big)^3 + 3 \times -\dfrac{3}{2} \\[1em] = -\dfrac{27}{8} - \dfrac{9}{2} \\[1em] = \dfrac{-27 - 36}{8} \\[1em] = \dfrac{-63}{8}.

Hence, (x3−1x3)=±638\Big(x^3 - \dfrac{1}{x^3}\Big) = \pm \dfrac{63}{8}.

Answered By

11 Likes

Related Questions