WMC 2018 Pursuit - Pack 2, C2

Algebra Level 2

Let

y = x + 1 x + 1 x + 1 . . . \large y=\sqrt{x}+\frac{1}{\sqrt{x}+\frac{1}{\sqrt{x}+\frac{1}{...}}}

If x = 12 x=12 , then y y can be written as a + b a+\sqrt{b} , find a + b a+b .

WMC Pursuit Questions


The answer is 5.

Julian Yu by Julian Yu , 19, Philippines

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1 solution

y = x + 1 x + 1 x + 1 ⋯ y = x + 1 y y 2 − x y − 1 = 0 y = x + x + 4 2 Note that the RHS > 0 and hence y > 0 = 12 + 12 + 4 2 = 2 + 3 \begin{aligned} y & = \sqrt x + \frac 1{\sqrt x+\frac 1{\sqrt x + \frac 1\cdots}} \\ y & = \sqrt x + \frac 1y \\ y^2 - \sqrt x y - 1 & = 0 \\ y & = \frac {\sqrt x + \sqrt{x + 4}}2 & \small \color{#3D99F6} \text{Note that the RHS }>0 \text{ and hence } y > 0 \\ & = \frac {\sqrt {12}+\sqrt{12+4}}2 \\ & = 2 + \sqrt 3 \end{aligned}

Therefore, a + b = 2 + 3 = 5 a+b = 2+3 = \boxed 5 .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir but we will get y = 12 ± 16 2 y = \dfrac{\sqrt{12} \pm \sqrt{16}}{2} then why we are only selecting the positive one ⟹ y = 12 + 16 2 \implies y = \dfrac{\sqrt{12} + \sqrt{16}}{2} .

Ram Mohith - 2 years, 8 months ago

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Because x > 0 \sqrt x > 0 and the R H S > 0 RHS >0 .

Chew-Seong Cheong - 2 years, 8 months ago

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