Solve for x x

Algebra Level 2

Solve for x x in the following expression

x = 6 + 6 + 6 + … x= \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{\dots}}}}

1 2.5 2 Can not be solved 3 6

Hana Wehbi by Hana Wehbi , 51, USA

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2 solutions

Hana Wehbi
Jul 6, 2017

The expression can be written as x = 6 + x ⟹ x 2 − x − 6 = 0 x=\sqrt{6+x}\implies x^2-x-6=0 by squaring both sides.

( x − 3 ) ( x + 2 ) = 0 ⟹ x = 3 (x-3)(x+2)=0 \implies x=3 , the other root is negative so we reject it.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

How do we know the limit defining this infinitely nested function converges?

Zach Abueg - 3 years, 11 months ago

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If you look at it like this 6 1 / 2 + 6 1 / 4 + 6 1 / 8 + ⋯ + 6 1 / 2 n 6^{1/2}+6^{1/4}+6^{1/8}+\dots+ 6^{1/2n} isn't that a geometric series whose ratio is less than one which indicates it converges.

Hana Wehbi - 3 years, 11 months ago

x = 6 + 6 + 6 + . . . x 2 = 6 + 6 + 6 + + . . . x 2 = 6 + x x 2 − x − 6 = 0 ( x − 3 ) ( x + 2 ) = 0 ⇔ x 1 = 3 ⇔ x 2 = − 2 x=\sqrt { 6+\sqrt { 6+\sqrt { 6+... } } } \\ { x }^{ 2 }=6+\sqrt { 6+\sqrt { 6+\sqrt { +... } } } \\ { x }^{ 2 }=6+x\\ { x }^{ 2 }-x-6=0\\ (x-3)(x+2)=0\\ \Leftrightarrow { x }_{ 1 }=3\\ \Leftrightarrow { x }_{ 2 }=-2

Thus, the value of x x that satisfies is x = 3 x = 3 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Thank you for sharing your solution.

Hana Wehbi - 3 years, 11 months ago

I think there is a typo when writing x 2 x_2 .

Hana Wehbi - 3 years, 11 months ago

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