If the expression x=sqrt(6+sqrt(6+sqrt(6+sqrt(6+...) extends to an

Question

If the expression  x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}  extends to an an infinite number of roots and converges to a positive number k, then approximately what is the value of k?

A. 1
B. 3
C. 5 + √5
D. 5 + √6
E. 5√6

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Regor60 · Answer

Squaring both sides yields

X^2 = 6 + X so

X^2 - X - 6 = 0

factoring the above yields

(X-3)(X+2) = 0

X=3 is the solution yielding a positive result

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av1901 · Answer

x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}

Squaring both sides is the best approach for such questions

x^2 = 6 + \sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}

Now since we know this goes on infinitely, we can say that the part after the 6 in the right hand side of the equation is equal to the original  x 
Rewrite the equation as

x^2 = 6 + x 
 x^2 - x - 6 = 0 
 x^2 + 2x - 3x - 6 = 0 
 (x+2)(x-3) = 0 
x = -2 or 3
Since x is a square root expression, it cannot be negative so
 x = 3  is the only solution

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gmatophobia · Answer

Given that

x =  \sqrt {6+\sqrt {6+\sqrt {6+..}}}

As the series tend to infinity, we can re-write the series as -

x =  \sqrt {6+x}

Squaring the equation on both sides -

x^2 = 6 + x

x^2 - x - 6 = 0

Factorizing the middle term

(x-3)(x+2) = 0

Therefore

x = 3 or x = -2

Note : RHS is a value that's under root, hence the result will always be positive.

Hence we can drop x = -2.

Ans :   x = 3 (Option : B)

tanyas311 · Answer

Given:  x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}

Squaring both sides gives us :  x^2=6+x  
=> x^2-x-6=0 
=> x^2-3x+2x-6=0 
=> (x+2)(x-3)=0 
=>x=-2,3

Since k is a positive integer, the value of x=3

Hence Answer is Option B

MissBong · Answer

Let x=sqrt(6+sqrt(6+sqrt(6+sqrt(6+...) 
Therefore,  x= \sqrt{6+x}

Squaring both sides,

x^2 = 6+x 
 x^2-x-6=0 
 (x-3)(x+2)=0 
Now, since the above equation can converge to a positive number  k , therefore,
 x=3

Therefore  k=3 
 Option B

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