If (2+(2+x)^(1/2))^(1/2) +(2-(2+x)^(1/2))^(1/2)=(2x)^(1/2), what is

Another way would be to understand that the square roots on left hand side are of form a+b and a-b.

Square both sides.
\((\sqrt{2+\sqrt{2+x}} +\sqrt{2-\sqrt{2+x}})^2=(\sqrt{2x})^2\)
\(2+\sqrt{2+x}+2-\sqrt{2+x}+2*\sqrt{2+\sqrt{2+x}} *\sqrt{2-\sqrt{2+x}}={2x}\)
\(4+2*\sqrt{(2^2-\sqrt{2+x}^2}=2x\)
\(4+2(\sqrt{2-x}=2x\)
\(\sqrt{2-x}=x-2\)
Square both sides again
\(2-x=(x-2)^2……(x-2)^2-(2-x)=0…..(x-2)^2+(x-2)=0……(x-2+1)(x-2)=0\)
Hence, x is 2 or -1.
But x cannot be negative. Hence, x is 2


E

Both 1 (choice C) and 2 (choice E) are correct


You can also solve this by calling y= root (2+x) and solving a quadratic equation that eventually comes through as x2 - 3x + 2= 0

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To cross check , plus in 1 and 2 in the equation and see

You require to be very careful while dealing with numbers and square roots.
I cannot comment on x^2-3x+2=0 as you have not mentioned how you got it.

Double check your calculations and you will find your mistake.

You can also put the value of x as 1.

\(\sqrt{2+\sqrt{2+x}} +\sqrt{2-\sqrt{2+x}}=\sqrt{2x}\)
\(\sqrt{2+\sqrt{2+1}} +\sqrt{2-\sqrt{2+1}}=\sqrt{2*1}\)
\(\sqrt{2+\sqrt{3}} +\sqrt{2-\sqrt{3}}=\sqrt{2}\)
\(\sqrt{2+1.732} +\sqrt{2-1.732}=\sqrt{2}\)
\(\sqrt{3.732} +\sqrt{0.268}}=\sqrt{2}\)
Is it possible??

You have TWO positive values getting added on left side and one of them on its own is greater than the right hand side

Hi,

Please check the following calculations:

\(\sqrt{2+\sqrt{2+x}} +\sqrt{2-\sqrt{2+x}}=\sqrt{2x}\)
Squaring both the sides, we get
\(4+2\sqrt{2-x}=2x\)
\(2+\sqrt{2-x}=x\)
\(\sqrt{2-x}=x-2\)
This will give us the equation x^2-3x+2=0, which is equal to (x-2)(x-1)=0.
Hence, the roots of the equation —> 2, 1.

How do we go about from here?

Do we need to check whether both the roots satisfy the equation?

Yes, it is important to check values.
As square root is positive but 1-2 makes it negative, 1 is not a valid value.

The caveat lies here, before proceeding further from this step, √(2-x)=(x-2) we need to check the domain of x. Also, the functions at both L.H.S and R.H.S should be greater than equal to 0, because L.H.S is the square root of some quantity and can never be less than 0.
for L.H.S √(2-x)>=0 , therefore x<=2.
for R.H.S (x-2)>=0, therefore x=>2.
The only value of x that will satisfy both L.H.S and R.H.S from x<=2 & x=>2 is x=2.

Alternatively, we can check.
√(2-x) = (x-2) [Take R.H.S to L.H.S]
√(2-x) -x+2 = 0
√(2-x) + (2-x) =0
√(2-x) + √(2-x)*√(2-x) =0 {(2-x) can be written as √(2-x)*√(2-x)}
√(2-x) (1+√(2-x)) = 0 [Take √(2-x) common]
Now product of two quantities is 0, hence we need to check for both,
√(2-x)=0 or (1+√(2-x))=0 { can never be 0, since it is (1+ positive quantity)}
Hence √(2-x)=0 which gives x=2.

Hope this helps.

Official Solution:


If \(\sqrt{2+\sqrt{2+x}} +\sqrt{2-\sqrt{2+x}}=\sqrt{2x}\), what is the value of \(x\) ?


A. \(-2\)
B. \(-\sqrt{2}\)
C. \(1\)
D. \(\sqrt{2}\)
E. \(2\)


APPROACH 1 - USING ANSWER OPTIONS AND ESTIMATION:

To succeed on the GMAT, it's essential to master working backward and estimation techniques. Let's see how we can apply these tools to this problem.

Firstly, note that since all numbers on the GMAT are real numbers, the even roots of negative numbers are undefined. Therefore, options A and B can be ruled out immediately because if \(x\) is negative, \(\sqrt{2x}\) will not be defined.

Let's check the other answer options.

If \(x = 1\), the right-hand side becomes \(\sqrt{2x}=\sqrt{2}\), while the left-hand side's first term will be \(\sqrt{2+\sqrt{2+1}} =\sqrt{2+\sqrt{3}}\). Since this expression is already greater than \(\sqrt{2}\), adding a positive value (\(\sqrt{2-\sqrt{2+x}}\)) to it will definitely result in a greater value than \(\sqrt{2}\). We can eliminate this option.

If \(x = \sqrt{2}\), the right-hand side becomes \(\sqrt{2x}=\sqrt{2*\sqrt{2}}\approx \sqrt{2*1.4}=\sqrt{2.8}\). Meanwhile, the left-hand side's first term will be \(\sqrt{2+\sqrt{2+x}} =\sqrt{2+\sqrt{2+\sqrt{2}}}\approx \sqrt{2+\sqrt{2+1.4}}= \sqrt{2+\sqrt{3.4}}\). Since \(\sqrt{3.4}\) is definitely greater than 1, we can conclude that \(\sqrt{2+\sqrt{3.4}} > \sqrt{2+1} > \sqrt{2.8}\). Therefore, adding some positive value (\(\sqrt{2-\sqrt{2+x}}\)) to this expression will definitely result in a greater value than \(\sqrt{2.8}\). We can eliminate this option as well.

Therefore, the answer must be E. Nonetheless, for practice sake, we can still test \(x = 2\):

The right-hand side becomes \(\sqrt{2x}=\sqrt{2*2}=2\), and the left-hand side becomes:

APPROACH 2 - ALGEBRA:

We can start by squaring both sides of the given equation:

Expanding the left-hand side and simplifying, we get:

Since the square root of a number cannot be negative, \(x-2\) must be positive or zero. This implies that \(x \geq 2\).

Squaring both sides of the equation \(\sqrt{2-x}=x-2\), we get:

\(x = 1\) or \(x = 2\). Since we found that \(x \geq 2\), the only valid solution is \(x = 2\).


Answer: E

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