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Mathematics

If x2+125x2=835x^2 + \dfrac{1}{25x^2} = 8\dfrac{3}{5}, find the values of:

(i) (x+15x)\Big(x + \dfrac{1}{5x}\Big)

(ii) (x3+1125x3)\Big(x^3 + \dfrac{1}{125x^3}\Big)

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Answer

(i) Given,

⇒x2+125x2=835⇒x2+125x2=435\Rightarrow x^2 + \dfrac{1}{25x^2} = 8\dfrac{3}{5} \\[1em] \Rightarrow x^2 + \dfrac{1}{25x^2} = \dfrac{43}{5}

We know that,

⇒(x+15x)2=x2+(15x)2+2×x×15x⇒(x+15x)2=x2+125x2+25⇒(x+15x)2=435+25⇒(x+15x)2=455⇒(x+15x)2=9⇒(x+15x)=9⇒(x+15x)=±3\Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = x^2 + \Big(\dfrac{1}{5x}\Big)^2 + 2 \times x \times \dfrac{1}{5x} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = x^2 + \dfrac{1}{25x^2} + \dfrac{2}{5} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = \dfrac{43}{5} + \dfrac{2}{5} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = \dfrac{45}{5} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = 9 \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big) = \pm 3 \\[1em]

Hence, (x+15x)=±3.\Big(x + \dfrac{1}{5x}\Big) = \pm 3.

(ii) We know that,

⇒(x+15x)3=(x)3+(15x)3+3×x×15x×(x+15x)⇒(x+15x)3=(x)3+(15x)3+35×(x+15x) ………(1)\Rightarrow \Big(x + \dfrac{1}{5x}\Big)^3 = (x)^3 + \Big(\dfrac{1}{5x}\Big)^3 + 3 \times x \times \dfrac{1}{5x} \times \Big(x + \dfrac{1}{5x}\Big) \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^3 = (x)^3 + \Big(\dfrac{1}{5x}\Big)^3 + \dfrac{3}{5} \times \Big(x + \dfrac{1}{5x}\Big) \text{ ………(1)}

Given,

(x+15x)=±3.\Big(x + \dfrac{1}{5x}\Big) = \pm 3.

Case 1:

(x+15x)=+3.\Big(x + \dfrac{1}{5x}\Big) = +3.

Substituting values we get :

⇒(3)3=x3+1125x3+35×(3)⇒27=x3+1125x3+95⇒x3+1125x3=27−95⇒x3+1125x3=135−95⇒x3+1125x3=1265⇒x3+1125x3=2515\Rightarrow (3)^3 = x^3 + \dfrac{1}{125x^3} + \dfrac{3}{5}\times(3) \\[1em] \Rightarrow 27 = x^3 + \dfrac{1}{125x^3} + \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = 27 - \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = \dfrac{135-9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = \dfrac{126}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = 25\dfrac{1}{5}

Case 2:

(x+15x)=−3.\Big(x + \dfrac{1}{5x}\Big) = -3.

Substituting values in equation (1), we get :

⇒(−3)3=x3+1125x3+35×(−3)⇒−27=x3+1125x3−95⇒x3+1125x3=−27+95⇒x3+1125x3=−135+95⇒x3+1125x3=−1265⇒x3+1125x3=−2515\Rightarrow (-3)^3 = x^3 + \dfrac{1}{125x^3} + \dfrac{3}{5}\times(-3) \\[1em] \Rightarrow -27 = x^3 + \dfrac{1}{125x^3} - \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = -27 + \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = \dfrac{-135 + 9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = -\dfrac{126}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = -25\dfrac{1}{5}

Hence, x3+125x3=±2515.x^3 + \dfrac{1}{25x^3} = \pm 25\dfrac{1}{5}.

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