Find the value!

Algebra Level 2

6 + 6 + 6 + 6 + 6 + ⋯ 2 2 2 2 2 = ? \large \sqrt[2]{6+\sqrt[2]{6+\sqrt[2]{6+\sqrt[2]{6+\sqrt[2]{6+\cdots}}}}} = \, ?

Hint : Try converting this expression as a quadratic equation.


The answer is 3.

Viki Zeta by Viki Zeta , 19, India

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2 solutions

Viki Zeta
Jul 5, 2016

Let x= 6 + 6 + 6 + 6 + 6 + . . . 2 2 2 2 2 ⟹ x 2 = 6 + 6 + 6 + 6 + 6 + 6 + . . . 2 2 2 2 2 ⟹ x 2 = 6 + x ⟹ x 2 − x − 6 = 0 ⟹ x 2 + 2 x − 3 x + 6 = 0 ⟹ x ( x + 2 ) − 3 ( x + 2 ) = 0 ⟹ ( x − 3 ) ( x + 2 ) = 0 ⟹ x = 3 , x = − 2 Here, on starting the expression we have an inequality x > 0 , so the answer is x = 3. As -2 is lesser than 0, it cannot be a solution satisfying the expression. This might me confusing, think a bit more. You have 6 2 . Which is 2.449... So on analyzing the solutions, the value cannot be negative. Therefore x = 3 is a solution. \text{Let x=}\sqrt[2]{6+\sqrt[2]{6+\sqrt[2]{6+\sqrt[2]{6+\sqrt[2]{6+...}}}}} \\ \implies x^2 = 6 +\sqrt[2]{6+\sqrt[2]{6+\sqrt[2]{6+\sqrt[2]{6+\sqrt[2]{6+...}}}}} \\ \implies x^2 = 6 + x \\ \implies x^2 - x - 6 = 0\\ \implies x^2 + 2x - 3x + 6 = 0 \\ \implies x(x+2) - 3(x+2) = 0 \\ \implies (x-3)(x+2) = 0\\ \implies x = 3, x = -2 \\ \text{Here, on starting the expression we have an inequality } x > 0 \text{, so the answer is x = 3. As -2 is lesser than 0,}\\\text{it cannot be a solution satisfying the expression.} \\ \text{This might me confusing, think a bit more. You have }\sqrt[2]{6} \\ \text{. Which is 2.449... So on analyzing the solutions, the value cannot be negative.}\\ \text{Therefore x = 3 is a solution.}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

First off, your method of proving that x = 3 x=3 is not accurate. Note that if x 2 = 9 , x = ± 3 ; x 2 = 4 , x = ± 2 x^2=9,\;x=\pm3;\;x^2=4,\;x=\pm2

I had a discussion with someone else on this matter before. Yes, the answer is 3 3 , but not for the reasons you stated. Refer to Calvin's comment in the solution to this question to know more.

Hung Woei Neoh - 4 years, 11 months ago

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Yeah. I'll make it clear wait.

Viki Zeta - 4 years, 11 months ago

What about now?

Viki Zeta - 4 years, 11 months ago
Trevor Arashiro
Jul 6, 2016

Let x= a ( a + 1 ) + a ( a + 1 ) + a ( a + 1 ) + a ( a + 1 ) + a ( a + 1 ) + . . . 2 2 2 2 2 \text{Let x=}\sqrt[2]{a(a+1)+\sqrt[2]{a(a+1)+\sqrt[2]{a(a+1)+\sqrt[2]{a(a+1)+\sqrt[2]{a(a+1)+...}}}}}

Then x = a + 1 x=a+1 .

Note that a number of the form a ( a + 1 ) a(a+1) is the only type of number for which the above is an integer.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Note that,

Let x= a ( a + 1 ) + a ( a + 1 ) + a ( a + 1 ) + a ( a + 1 ) + a ( a + 1 ) + . . . 2 2 2 2 2 ⟹ x 2 = a ( a + 1 ) + x ⟹ x 2 − x − a ( a + 1 ) = 0 ⟹ x 2 − x − a 2 − a = 0 ⟹ ( x + a ) ( x − a ) − ( x + a ) = 0 ⟹ ( x + a ) ( x − a − 1 ) = 0 ⟹ x = − a , x = a + 1 So that, -2 is also a solution, but you must explain why only a+1 is a solution. \text{Let x=}\sqrt[2]{a(a+1)+\sqrt[2]{a(a+1)+\sqrt[2]{a(a+1)+\sqrt[2]{a(a+1)+\sqrt[2]{a(a+1)+...}}}}} \\ \implies x^2 = a(a+1) + x \\ \implies x^2 - x - a(a+1) = 0 \\ \implies x^2 - x - a^2 - a = 0 \\ \implies (x+a)(x-a) - (x+a) = 0 \\ \implies (x+a)(x-a-1) = 0 \\ \implies x = -a, x = a + 1 \\ \text{So that, -2 is also a solution, but you must explain why only a+1 is a solution.}

Viki Zeta - 4 years, 11 months ago

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