Not an irrational

Algebra Level 3

Find x x .

x = 4 + 3 + 6 ( 4 + 3 + 6 ( 4 + 3 + 6 ( 4 + . . . ) ) ) x=4+\sqrt{3+6\left(4+\sqrt{3+6\left(4+\sqrt{3+6(4+...)}\right)}\right)}

14 12 13 11

Pnn Sumedh by Pnn Sumedh , 21, India

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2 solutions

Chew-Seong Cheong
Aug 11, 2017

x = 4 + 3 + 6 ( 4 + 3 + 6 ( 4 + 3 + 6 ( 4 + . . . ) ) ) x = 4 + 3 + 6 x x − 4 = 3 + 6 x Squaring both sides x 2 − 8 x + 16 = 3 + 6 x x 2 − 14 x + 13 = 0 ( x − 1 ) ( x − 13 ) = 0 ⟹ x = 13 Since x > 4 \begin{aligned} x & = 4 + \sqrt{3+6\left(4 + \sqrt{3+6\left(4 + \sqrt{3+6\left(4 + ...\right)} \right)} \right)} \\ x & = 4 + \sqrt{3+6x} \\ x - 4 & = \sqrt {3+6x} & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 - 8x + 16 & = 3 + 6x \\ x^2 - 14x + 13 & = 0 \\ (x-1)(x-13) & = 0 \\ \implies x & = \boxed{13} & \small \color{#3D99F6} \text{Since }x >4 \end{aligned}

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Mahdi Raza
Jun 8, 2020

x = 4 + 3 + 6 x ( x − 4 ) 2 = 3 + 6 x x 2 − 8 x + 16 − 3 + 6 x = 0 ( x − 1 ) ( x − 13 ) = 0 ⟹ x = 13 [ x > 4 ] \begin{aligned} x &= 4 + \sqrt{3 + 6x} \\ (x-4)^2 &= 3+6x \\ x^2 - 8x + 16 - 3 + 6x &= 0 \\ (x-1)(x-13) &= 0 \\ \implies x &= 13 &[x>4] \end{aligned}

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