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Algebra Level 3

Find x ≠ 0 x\ne 0 that satisfies the following: 1 1 1 1 x + 1 2 + 1 1 x + 1 2 + 1 1 1 x + 1 2 + 1 1 x + 1 2 = x 36 \large \dfrac{1}{\dfrac{1}{\dfrac{1}{\dfrac{1}{x}+\dfrac{1}{2}}+\dfrac{1}{\dfrac{1}{x}+\dfrac{1}{2}}}+\dfrac{1}{\dfrac{1}{\dfrac{1}{x}+\dfrac{1}{2}}+\dfrac{1}{\dfrac{1}{x}+\dfrac{1}{2}}}} =\dfrac{x}{36}

70 80 81 71 100

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Chew-Seong Cheong
Jul 31, 2017

x 36 = 1 1 1 1 x + 1 2 + 1 1 x + 1 2 + 1 1 1 x + 1 2 + 1 1 x + 1 2 = 1 2 2 1 x + 1 2 = 1 2 2 2 + x 2 x = 1 2 4 x 2 + x = 1 4 + 2 x 4 x \large \begin{aligned} \frac x{36} & = \frac 1{\frac 1{\frac 1{\frac 1x+\frac 12} +\frac 1{\frac 1x+\frac 12}} + \frac 1{\frac 1{\frac 1x+\frac 12} +\frac 1{\frac 1x+\frac 12}}} \\ & = \frac 1{\frac 2{\frac 2{\frac 1x+\frac 12}}} = \frac 1{\frac 2{\frac 2{\frac {2+x}{2x}}}} = \frac 1{\frac 2{\frac {4x}{2+x}}} = \frac 1{\frac {4+2x}{4x}} \end{aligned}

⟹ x 36 = 4 x 4 + 2 x 2 x 2 + 4 x = 144 x x ( x − 70 ) = 0 ⟹ x = 70 Since x ≠ 0 \begin{aligned} \implies \frac x{36} & = \frac {4x}{4+2x} \\ 2x^2 + 4x & = 144 x \\ x(x-70) & = 0 \\ \implies x & = \boxed{70} & \small \color{#3D99F6} \text{Since }x \ne 0 \end{aligned}

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@Hana Nakkache , you have to mentioned that "for x ≠ 0 x \ne 0 ", else the equation is undefined.

Chew-Seong Cheong - 3 years, 10 months ago

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Ok, l will. Thank you for the solution and remark.

Hana Wehbi - 3 years, 10 months ago

No, that's not necessary. If x=0, then LHS is undefined, and RHS = 0, so x=0 cannot be a solution

Pi Han Goh - 3 years, 10 months ago

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