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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 13 Aug 2023, 06:00
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The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4­

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D
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 13 Aug 2023, 15:06
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The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4
Show more

On the set of real numbers, square roots are defined only if the expressions under the square roots are non-negative.

x + 2 ≥ 0
x ≥ -2

4 – x ≥ 0
4 ≥ x

√(x + 2) - √(4 – x) ≥ 0
√(x + 2) ≥ √(4 – x)
x + 2 ≥ 4 – x
2x ≥ 2
x ≥ 1

Since x ≥ -2 AND 4 ≥ x AND x ≥ 1, we have:

1 ≤ x ≤ 4

Answer: D
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 13 Aug 2023, 06:12
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Bunuel
The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4
Show more

The value of an expression under root is always non - negative

  • \(\sqrt{x+2}\)

    \(x + 2 \geq 0\)

    \(x \geq -2\)

  • \(\sqrt{4-x}\)

    \(4-x \geq 0\)

    \(x \leq 4\)

  • \(\sqrt{\sqrt{x+2} - \sqrt{4-x} } \)

    \(\sqrt{x+2} - \sqrt{4-x} \geq 0 \)

    \(\sqrt{x+2} \geq \sqrt{4-x}\)

    Applying the principle that \(\sqrt{n} = |n|\)

    \(|x+2| \geq |4-x|\)

    Squaring both sides we get

    \(x^2 + 4x + 4 \geq 16 + x^2 - 8x\)

    \(x^2 + 4x + 8x - x ^2 \geq 16 - 4 \)

    \(12x \geq 12 \)

    \(x \geq 1 \)

Combining the information we get -

\(1 \leq x \leq 4\)

Option D
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 05 Feb 2024, 00:39
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gmatophobia

Bunuel
The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4
The value of an expression under root is always non - negative


  • \(\sqrt{x+2}\)

    \(x + 2 \geq 0\)

    \(x \geq -2\)
  • \(\sqrt{4-x}\)

    \(4-x \geq 0\)

    \(x \leq 4\)
  • \(\sqrt{\sqrt{x+2} - \sqrt{4-x} } \)

    \(\sqrt{x+2} - \sqrt{4-x} \geq 0 \)

    \(\sqrt{x+2} \geq \sqrt{4-x}\)

    Applying the principle that \(\sqrt{n} = |n|\)

    \(|x+2| \geq |4-x|\)

    Squaring both sides we get

    \(x^2 + 4x + 4 \geq 16 + x^2 - 8x\)

    \(x^2 + 4x + 8x - x ^2 \geq 16 - 4 \)

    \(12x \geq 12 \)

    \(x \geq 1 \)

Combining the information we get -

\(1 \leq x \leq 4\)

Option D




I completely get the logics but can I understand from you why \(x \geq -2\), \(x \leq 4\), \(x \geq 1 \) at the end it is \(1 \leq x \leq 4\), why not \(-2 \leq x \leq 4\)?





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The conditions (\(x \geq -2\), \(x \leq 4\), and \(x \geq 1\)) represent restrictive conditions, meaning for the whole expression to be defined they must be true simultaneously. You see, if \(-2 \leq x < 1\), then \(\sqrt{\sqrt{x + 2} - \sqrt{4 – x}} < 0\), making this expression undefined for real numbers in that range.­
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 11 Jan 2024, 21:15
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gmatphobia chetan2u Bunuel, how do we figure that we have to equate expressions as well?
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 12 Jan 2024, 00:23
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gmatphobia chetan2u Bunuel, how do we figure that we have to equate expressions as well?
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That's because the even roots are defined for values that are more than or equal to zero.
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 14 Jan 2024, 12:07
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Bunuel
The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4
Show more

We can PLUG IN THE ANSWERS, which represent the domain of the function.

Options A and B include x=0; the remaining options do not.
If x=0, we get:
\(f(0) = \sqrt{\sqrt{2} - \sqrt{4} } ≈ \sqrt{1.4 - 2} = \sqrt{-0.6} \)
Not viable, since the value under the root symbol may not be negative.
Since x=0 is not viable, eliminate A and B.

Options D and E include x=4; option C does not.
If x=4, we get:
\(f(4) = \sqrt{\sqrt{6} - \sqrt{0} } ≈ \sqrt{\sqrt{6}} \)
This works.
Eliminate C, which does not include x=4.

Options D includes x=1; option E does not.
If x=1, we get:
\(f(1) = \sqrt{\sqrt{3} - \sqrt{3} } = \sqrt{0} \)
This works.
Eliminate E, which does not include x=1.

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D
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 04 Feb 2024, 15:30
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Bunuel
The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4

The value of an expression under root is always non - negative

  • \(\sqrt{x+2}\)

    \(x + 2 \geq 0\)

    \(x \geq -2\)
  • \(\sqrt{4-x}\)

    \(4-x \geq 0\)

    \(x \leq 4\)
  • \(\sqrt{\sqrt{x+2} - \sqrt{4-x} } \)

    \(\sqrt{x+2} - \sqrt{4-x} \geq 0 \)

    \(\sqrt{x+2} \geq \sqrt{4-x}\)

    Applying the principle that \(\sqrt{n} = |n|\)

    \(|x+2| \geq |4-x|\)

    Squaring both sides we get

    \(x^2 + 4x + 4 \geq 16 + x^2 - 8x\)

    \(x^2 + 4x + 8x - x ^2 \geq 16 - 4 \)

    \(12x \geq 12 \)

    \(x \geq 1 \)

Combining the information we get -

\(1 \leq x \leq 4\)

Option D
 
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I completely get the logics but can I understand from you why \(x \geq -2\), \(x \leq 4\), \(x \geq 1 \) at the end it is \(1 \leq x \leq 4\), why not  \(-2 \leq x \leq 4\)?
 
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 22 Jul 2024, 10:01
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would plugging in the values help here?
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would plugging in the values help here?
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­difficult to get roots particularly ­combined roots with plugging in and time constraints
\(\sqrt{x+2} - \sqrt{4-x}\)
­
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 16 Dec 2024, 07:38
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Solved the question using options.

Here is the detailed explanation



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Bunuel
The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4­

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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 06 Jan 2025, 04:03
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Bunuel
The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4­

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Responding to a pm:

Domain means the values of x for which the expression will take real values. For example, we cannot put x = -5 because then (x+2), which is under a square root, will become negative. That is not allowed since we cannot find the square root of negative numbers.
Usually, when looking for domain, the common constraints are that the square root should not be negative and the denominator should not be 0.

There are multiple ways of solving this question.

Method 1: Ensure all square roots are non negative so we need to ensure that x + 2 >= 0, 4 - x >= 0 and \(\sqrt{x+2} - \sqrt{4-x} \geq 0\)
Show by solutions above.

Method 2: Plug in Values
I would prefer this since it is very quick. Simplest would be to check for x = 0, 3 and 1
Every value of x for which f(x) has a real value is a part of the domain.

When we put x = 0, we see that f(x) is not real and hence option (A) and (B) cannot be the domain since as per them, x can take value 0.
When we put x = 3, we see that f(x) is real so x = 3 is possible. So eliminate option (C) because it does not include 3.
When we put x = 1, we see that f(x) is real so eliminate option (E) because it does not include 1.

Answer (D)
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How do we know that the numbers under the square roots must >= 0? The question says that the domain of x is a set of real numbers, but does not say the range will be real (i.e. that numbers under the square root must be positive & squart root value must be real)?
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 15 Jan 2025, 13:41
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GMAT does not test on imaginary or complex numbers, hence it is fair to assume that values underneath the square root are greater than or equal to 0.

Also think, if the problem allowed complex number in range couldn't domain include any value of x, even those not mentioned in the answer choices.
studybuggy
How do we know that the numbers under the square roots must >= 0? The question says that the domain of x is a set of real numbers, but does not say the range will be real (i.e. that numbers under the square root must be positive & squart root value must be real)?
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 02 Jun 2025, 13:32
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hi gmatophobia.

you said:

sqrt(n)=|n|

is this correct? I thought only this concept applies: sqrt(n ^ 2)=|n|. So just when the variable is squared then the square of that is equal to mod of the variable. Could you please clarify? thanks



gmatophobia
Bunuel
The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4

The value of an expression under root is always non - negative

  • \(\sqrt{x+2}\)

    \(x + 2 \geq 0\)

    \(x \geq -2\)
  • \(\sqrt{4-x}\)

    \(4-x \geq 0\)

    \(x \leq 4\)

  • \(\sqrt{\sqrt{x+2} - \sqrt{4-x} } \)

    \(\sqrt{x+2} - \sqrt{4-x} \geq 0 \)

    \(\sqrt{x+2} \geq \sqrt{4-x}\)

    Applying the principle that \(\sqrt{n} = |n|\)

    \(|x+2| \geq |4-x|\)

    Squaring both sides we get

    \(x^2 + 4x + 4 \geq 16 + x^2 - 8x\)

    \(x^2 + 4x + 8x - x ^2 \geq 16 - 4 \)

    \(12x \geq 12 \)

    \(x \geq 1 \)

Combining the information we get -

\(1 \leq x \leq 4\)

Option D
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 03 Jun 2025, 21:23
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gmatophobia did not say this. They said "The value of an expression under root is always non - negative"
i.e. if we have \(\sqrt{n}\) then n must be a positive number. GMAT does not include the concept of complex numbers hence any square root you see in GMAT implies that the term under the root will be non negative. Only then can you find its root.
\(\sqrt{-4}\) is not included in GMAT.
Also keep in mind that \(\sqrt{4} = 2\) only. It is not 2 or -2. Here are these and some other rules: https://anaprep.com/algebra-squares-and-square-roots/

Also, you can tag somebody using @ before their name. They will get a notification of their mention in that case.


jdtg2610
hi gmatophobia.

you said:

sqrt(n)=|n|

is this correct? I thought only this concept applies: \(sqrt(n ^ 2)=|n|\). So just when the variable is squared then the square of that is equal to mod of the variable. Could you please clarify? thanks



gmatophobia
Bunuel
The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4

The value of an expression under root is always non - negative

  • \(\sqrt{x+2}\)

    \(x + 2 \geq 0\)

    \(x \geq -2\)
  • \(\sqrt{4-x}\)

    \(4-x \geq 0\)

    \(x \leq 4\)

  • \(\sqrt{\sqrt{x+2} - \sqrt{4-x} } \)

    \(\sqrt{x+2} - \sqrt{4-x} \geq 0 \)

    \(\sqrt{x+2} \geq \sqrt{4-x}\)

    Applying the principle that \(\sqrt{n} = |n|\)

    \(|x+2| \geq |4-x|\)

    Squaring both sides we get

    \(x^2 + 4x + 4 \geq 16 + x^2 - 8x\)

    \(x^2 + 4x + 8x - x ^2 \geq 16 - 4 \)

    \(12x \geq 12 \)

    \(x \geq 1 \)

Combining the information we get -

\(1 \leq x \leq 4\)

Option D
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Bunuel
The domain of the function \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x} }\) is the set of real numbers x such that

A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤ 4

The value of an expression under root is always non - negative

  • \(\sqrt{x+2}\)

    \(x + 2 \geq 0\)

    \(x \geq -2\)
  • \(\sqrt{4-x}\)

    \(4-x \geq 0\)

    \(x \leq 4\)

  • \(\sqrt{\sqrt{x+2} - \sqrt{4-x} } \)

    \(\sqrt{x+2} - \sqrt{4-x} \geq 0 \)

    \(\sqrt{x+2} \geq \sqrt{4-x}\)

    Applying the principle that \(\sqrt{n} = |n|\)

    \(|x+2| \geq |4-x|\)

    Squaring both sides we get

    \(x^2 + 4x + 4 \geq 16 + x^2 - 8x\)

    \(x^2 + 4x + 8x - x ^2 \geq 16 - 4 \)

    \(12x \geq 12 \)

    \(x \geq 1 \)

Combining the information we get -

\(1 \leq x \leq 4\)

Option D
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you have applied the wrong principle in the highlighted part it is \sqrt{n^2} when the term coverts to mod and becomes |n|
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The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x 08 Oct 2025, 06:08
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Let me walk you through the key thinking process here.

Understanding What We're Dealing With

You have \(f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x}}\), and you need to find where this function is actually defined. Here's the thing: this isn't just about making sure the inner square roots work - you also need to make sure that outer square root has something non-negative inside it. That's where most students trip up.

Step 1: Check the Inner Square Root Constraints

Let's start with the basics. For \(\sqrt{x+2}\) to be defined, you need:
\(x + 2 \geq 0\)
\(x \geq -2\)

For \(\sqrt{4-x}\) to be defined, you need:
\(4 - x \geq 0\)
\(x \leq 4\)

So far, combining these gives you \(-2 \leq x \leq 4\). But notice - this would be answer choice B, and if you selected that, you'd be missing a crucial constraint.

Step 2: The Key Insight - The Outer Square Root Constraint

Here's what you need to see: the outer square root means that the entire expression \(\sqrt{x+2} - \sqrt{4-x}\) must be non-negative. In other words:

\(\sqrt{x+2} - \sqrt{4-x} \geq 0\)

This means:
\(\sqrt{x+2} \geq \sqrt{4-x}\)

Think about what this is telling you: the first square root has to be at least as large as the second one. This creates an additional restriction on x that most students completely miss!

Step 3: Solve the Inequality

Since both sides are non-negative (they're square roots), you can square both sides:

\(x + 2 \geq 4 - x\)
\(x + x \geq 4 - 2\)
\(2x \geq 2\)
\(x \geq 1\)

Step 4: Combine All Constraints

Now let's put everything together:
  • From \(\sqrt{x+2}\): \(x \geq -2\)
  • From \(\sqrt{4-x}\): \(x \leq 4\)
  • From the outer square root: \(x \geq 1\)

The most restrictive lower bound is \(x \geq 1\), and the upper bound is \(x \leq 4\).

Final Answer: \(1 \leq x \leq 4\) (Choice D)

You can verify this makes sense: at \(x = 1\), you get \(f(1) = \sqrt{\sqrt{3} - \sqrt{3}} = 0\), which works perfectly. At \(x = 4\), you get \(f(4) = \sqrt{\sqrt{6} - 0} = \sqrt[4]{6}\), which is defined.

The Takeaway

The critical error that leads students to pick B is missing that third constraint. When you have nested radicals, you need to ensure that every single layer is properly defined - not just the innermost expressions.

For a complete understanding of the systematic approach to domain problems with nested functions, including how to spot all constraint types and avoid common algebraic errors, you can check out the detailed solution on Neuron by e-GMAT. You'll also find the framework that applies to all nested function problems, plus pattern recognition techniques to save time. Feel free to explore comprehensive solutions for other official GMAT questions on Neuron to build systematic accuracy across question types.

Hope this helps clarify the logic! Let me know if you have questions about any of the steps.
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