If the expression x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...) extends to an

Question

If the expression  x=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}  extends to an infinite number of roots and converges to a positive number x, what is x?

A.  \sqrt 3

B. 2

C.  1+\sqrt 2

D.  1+\sqrt 3

E.  2*\sqrt 3

Bunuel · Accepted Answer

x=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}  -->  x=\sqrt{2+({\sqrt{2+\sqrt{2+\sqrt{2+...})}}}} , as the expression under square root extends infinitely then expression in brackets would equal to  x  itself so we can rewrite given expression as  x=\sqrt{2+x} . Square both sides  x^2=2+x  -->  x=2  or  x=-1 . As given that  x>1  then only one solution is valid:  x=2 .

Answer: B.

mikemcgarry · Answer

Dear mun23 , The trick of this question is to give the entire expression a name --- I am going to call it S, and then I am going to square it. nested roots.JPG Square the expression produces 2 plus a copy of itself --- that's why we can replace it on the other side with S, and then solve for S algebraically: S^2 = 2 + S S^2 - S - 2 = 0 (S - 2)(S + 1) = 1 S = 2 or S = -1 The negative root makes no sense in this context, so S = 2, and the answer = B Does all this make sense? Mike :-)

IanStewart · Answer

This question is out of scope for the GMAT, but there's an interesting trick to questions like this. We know that:

x=sqrt(2+ sqrt(2+sqrt(2+sqrt(2+... )

Now, notice that the part I've highlighted in red is actually equal to x itself. So we can replace it with x, to get the much simpler equation:

x = sqrt(2 + x)
x^2 = 2 + x
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 or -1

and since x cannot be negative, x = 2. You won't see anything like this on the GMAT though, so it's for interest only.

trunksli · Answer

That's a clever solution, putting x into the equation itself. Sort of recursive, and very elegant. 
These type of problems give me a hard time.

mun23 · Answer

Hi mikemcgarry
I am not understanding why the entire expression is given a named?

mikemcgarry · Answer

Dear mun23 , This is a standard trick in mathematics. We give the entire expression a name, the name S, a variable, because that allows us to manipulate it algebraically. We want to know the value of the entire expression, so we set the entire expression equal to a variable, then ultimately all we have to do is solve for the value of this variable. Because the variable equals the whole expression, when we know the value of the variable, we also know the value of the whole expression. This is an extension of the fundamental power of algebra --- when we assign a variable to any unknown quantity, then the whole panoply of algebraic techniques comes to bear on the problem. Keep in mind that material like this ---- infinitely recursive expressions --- is exceedingly unlike to appear on the GMAT. I have never seen anything like this. If it did appear at all, it would only appear to someone getting virtually everything else correct on the Quant section. Folks in the Q < 45 range will NEVER see a question about this stuff, and even folks in the high 50s would only see it less than 1% of the time. Does all this make sense? Mike :-)

Bunuel · Answer

Merging similar topics. Please refer to the solutions above. Similar questions to practice: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029228 find-the-value-of-a-given-a-3-3-3-3-3-inf-138049.html find-the-value-of-x-75403.html Hope it helps.

AccipiterQ · Answer

Shouldn't the answer to this be infinity....I have been looking at this one for about 45 minutes, and I can't figure it out. We start with 2, and then add to that  \sqrt{2} , which is about 1.4, then we add to that the square root of the square root of 2, or about 1.18, and then add the square root of the square root of the square root of 2, which is 1.09. The numbers CAN NOT ever be below 1. Just taking it out through 10 cycles, the total is almost 11. And this is an infinite sequence, so the answer is whatever the square root of infinity is. I've looked at the solutions above and they don't make sense to me, at all. The way the problem is written, the answer CANT be any of the options listed

Bunuel · Answer

Consider the examples below:
 \sqrt{2}\approx{1.4} ;

\sqrt{2+\sqrt{2}}\approx{1.85} ;

\sqrt{2+{\sqrt{2+\sqrt{2}}}}\approx{1.96} ;

\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2}}}}}\approx{1.99} ;

\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}\approx{1.998} ;

\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}}\approx{1.9994} .

...

As you can see the result approaches to 2 by decreasing pace. If we extend that to an infinite number of roots the result will be exactly 2.

SVaidyaraman · Answer

Main Idea: Identifying that the problem can be solved by expressing what is given, by an equation. Since there is only one equation possible and if LHS is x , the RHS should also contain x.

Details: We see the given info can be expressed as x=sqrt(2+x). So we can now solve for x . X turns out to be 2 and -1. We reject the negative value and take x=2

Hence B.

IanStewart · Answer

I fear this might give test takers the wrong impression. If you are doing very well on the GMAT, and you are thus seeing the hardest questions, it is not true that you start seeing questions testing obscure or advanced math knowledge (like how to compute infinite sums). You see questions testing the same math you're tested on at the Q40 level -- ratios, divisibility, basic algebra -- but the questions will test those core concepts in trickier or more logically intricate ways.

You can simply never see a question like the one in this thread on the GMAT, because to prove an infinite sum "converges" at all (does not run off to infinity), you need calculus, and calculus is not within the scope of the test.

CrackverbalGMAT · Answer

On squaring on both sides,
 X^2 = 2+ X 
 X^2 - X -2 =0 
 ( X- 2 )( X+ 1 )= 0

X = 2,-1

Since X is positive , then we can say X = 2.

Option B is the answer.

Thanks,
 Clifin J Francis,
GMAT SME

djangobackend · Answer

x>0 and not x>1

Bunuel · Answer

The condition is x > 1 because x equals √(2 + something), and since that "something" is a nested positive expression, the entire value inside the square root is clearly more than 2. So √(2 + something) must be greater than √2, which is already more than 1. Hence, x > 1 is correct.

velvetwhisper · Answer

I used the shortcut formula for this question. 
the formula is {⎷(4a+1) +1}/2
Now substitute 2 in place of "a", so the equation will be:
{⎷(4x2+1) +1}/2
={⎷(8+1) +1}/2
={⎷9 +1}/2
={3+1}/2 = 2

If the expression x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...) extends to an

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If the expression x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...) extends to an

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