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805+ Level|   Algebra|      
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gmatophobia
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­If c is a nonnegative constant, how many real number values for x sat 17 Feb 2024, 07:59
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C
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95% (hard)

Question Stats:

43% (01:41) correct 57% (01:46) wrong based on 565 sessions

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­If c is a nonnegative constant, how many real number values for x satisfy the equation \(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

A. 0

B. 1

C. 2

D. 3

E. Infinitely many­­

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C
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Bunuel
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­If c is a nonnegative constant, how many real number values for x sat 17 Feb 2024, 08:22
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gmatophobia
­If c is a nonnegative constant, how many real number values for x satisfy the equation \(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

A. 0

B. 1

C. 2

D. 3

E. Infinitely many­

Squaring the given expression, we obtain:

    ­\(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

    \(c - x =c - 2\sqrt{cx} + x\)

    \(x = \sqrt{cx}\)
­­­

Now, square again:

    \(x^2 = cx\)

    \(x^2 - cx = 0\)

    \(x(x - c) = 0\)
­­­­­

The above equation has two solutions, x = 0 and x = c.

Answer: C.­­
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­If c is a nonnegative constant, how many real number values for x sat 17 Feb 2024, 22:33
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Bunuel
 
gmatophobia
­If c is a nonnegative constant, how many real number values for x satisfy the equation \(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

A. 0

B. 1

C. 2

D. 3

E. Infinitely many­

Squaring the given expression, we obtain:

    ­\(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

    \(c - x =c - 2\sqrt{cx} + x\)

    \(x = \sqrt{cx}\)
­­­

Now, square again:

    \(x^2 = cx\)

    \(x^2 - cx = 0\)

    \(x(x - c) = 0\)
­­­­­

The above equation has two solutions, x = 0 and x = c.

Answer: C.­­
­Hi Bunuel, 
As x = c so any nonnegative value of c is value of x. Thu,
Should not the answer be E then?
 
Thanks.
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­If c is a nonnegative constant, how many real number values for x sat 17 Feb 2024, 22:45
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unraveled
Bunuel
 
gmatophobia
­If c is a nonnegative constant, how many real number values for x satisfy the equation \(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

A. 0

B. 1

C. 2

D. 3

E. Infinitely many­

Squaring the given expression, we obtain:

    ­\(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

    \(c - x =c - 2\sqrt{cx} + x\)

    \(x = \sqrt{cx}\)
­­­

Now, square again:

    \(x^2 = cx\)

    \(x^2 - cx = 0\)

    \(x(x - c) = 0\)
­­­­­

The above equation has two solutions, x = 0 and x = c.

Answer: C.­­
­Hi Bunuel, 
As x = c so any nonnegative value of c is value of x. Thu,
Should not the answer be E then?
 
Thanks.
­
Ans is B 2

c has a finite fixed value choosen from a infinite range ( it may take any value between 0 to Inf ) , but once the constant is decided like let's say c = 3

then the equation

\(\sqrt{(c - x)}  =  \sqrt{c} - \sqrt{x}\)
\(\sqrt{(3 - x)}  =  \sqrt{3} - \sqrt{x}\)

will have only two values as root.

\(c - x = c + x - 2\sqrt{(c*x)}\)  only when \(x = \sqrt{(c*x)}\)
that is possible only when x = c or x = 0  , here value of x will be 0 or 3.­­­
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­If c is a nonnegative constant, how many real number values for x sat 18 Feb 2024, 03:08
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x=0 and x=c are 2 solutions.­­
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unraveled
Bunuel
 
gmatophobia
­If c is a nonnegative constant, how many real number values for x satisfy the equation \(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

A. 0

B. 1

C. 2

D. 3

E. Infinitely many­

Squaring the given expression, we obtain:

    ­\(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

    \(c - x =c - 2\sqrt{cx} + x\)

    \(x = \sqrt{cx}\)
­­­

Now, square again:

    \(x^2 = cx\)

    \(x^2 - cx = 0\)

    \(x(x - c) = 0\)
­­­­­

The above equation has two solutions, x = 0 and x = c.

Answer: C.­­
­Hi Bunuel, 
As x = c so any nonnegative value of c is value of x. Thu,
Should not the answer be E then?
 
Thanks.
­c there is a fixed constant, not a variable. So it has one fixed value, say 10 or 12.
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Bunuel
unraveled
Squaring the given expression, we obtain:

    ­\(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

    \(c - x =c - 2\sqrt{cx} + x\)

    \(x = \sqrt{cx}\)
­­­

Now, square again:

    \(x^2 = cx\)

    \(x^2 - cx = 0\)

    \(x(x - c) = 0\)
­­­­­

The above equation has two solutions, x = 0 and x = c.

Answer: C.­­
­Hi Bunuel, 
As x = c so any nonnegative value of c is value of x. Thu,
Should not the answer be E then?
 
Thanks.
­c there is a fixed constant, not a variable. So it has one fixed value, say 10 or 12.[/quote]
­yes, i realised it later after a couple of minutes of posting the question. Silly me. 
Only that if someone tries testing numbers on hit and trial basis, he/she is going to get wrong answer.­
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­If c is a nonnegative constant, how many real number values for x sat 03 Oct 2024, 00:27
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Since Zero is also nonnegative, When C = 0, X has only 1 possible solution and in other cases it has 2 solutions. Isn't the question ambiguous?
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Since Zero is also nonnegative, When C = 0, X has only 1 possible solution and in other cases it has 2 solutions. Isn't the question ambiguous?

Yes, in the general case, the equation has two solutions for x. However, in the specific case where c = 0, it has only one solution, x = 0.
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KrishnaVamsiP
Since Zero is also nonnegative, When C = 0, X has only 1 possible solution and in other cases it has 2 solutions. Isn't the question ambiguous?
To this extent it is surely ambiguous.
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­If c is a nonnegative constant, how many real number values for x sat 22 Dec 2024, 09:34
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Yes this question is a test question in the practice test of gmat official mock tests
I got the question wrong on the test but still got a Q90 in the mock
Blackcrow1972
KrishnaVamsiP
Since Zero is also nonnegative, When C = 0, X has only 1 possible solution and in other cases it has 2 solutions. Isn't the question ambiguous?
To this extent it is surely ambiguous.
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­If c is a nonnegative constant, how many real number values for x sat 11 Jan 2025, 00:24
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gmatophobia
­If c is a nonnegative constant, how many real number values for x satisfy the equation \(\sqrt{c - x} = \sqrt{c} - \sqrt{x}\)

A. 0

B. 1

C. 2

D. 3

E. Infinitely many­

Show Spoiler
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Screenshot_2024-02-17-20-26-10-06_b72a20be883aec8a014bd2b7c7038e87.jpg
­

Yes, there is a problem here. They needed to give that "c takes a positive constant value" (not non negative). Then the answer will be a clean 2.
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