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From statement I alone, √x=x-2.

Squaring both sides, we have x = \((x-2)^2\). Opening up the RHS and simplifying, we have,

\(x^2\) – 5x + 4 = 0, which can be factorized as (x-4) (x-1) = 0. Therefore, x has two values, 1 and 4. We need a unique value of x. Statement I alone is insufficient.

Answer options can be B, C or E. Answer options A and D can be eliminated.

From statement II alone, x≠1. This is not sufficient to find the value of x.
Answer option B can be eliminated.

Combining statements I and II, we have the following:

From statement I, x = 1 or 4. From statement II, x≠1.
This means, x = 4. The combination of statements is sufficient. Answer option E can be eliminated.

The correct answer option is C.

Hope that helps!
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solve for #1
we get
x=(x-2)^2
or say
x^2-5x+4=0
so x= 4 or 1
but since LHS is +ve so 1 is not possible at square root value at RHS
x=4 sufficient
#2
x ≠ 1
insufficient
Option A

What is the value of x?

(1) x√=x−2

(2) x ≠ 1
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Quote:

What is the value of x?

(1) √x=x−2

(2) x ≠ 1

(1) √x=x−2 sufic

|x|=(x-2)^2
|x|=x^2+4-4x
x≥0: x=x^2+4-4x, x^2-5x-4=0, (x-4)(x-1)=0; x=4,1
x=1: √1=1-2, 1=-1=invalid
x=4: √4=4-2, 2=2=valid

(2) x ≠ 1 insufic

Ans (A)
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What is the value of x?

(1) √x = x−2
\(x = x^2 + 4 - 4x\)
\(x^2 + 4 - 5x = 0\)
(x-4)(x-1) = 0
x = 4 OR 1

INSUFFICIENT.

(2) x ≠ 1
Any value is possible.

INSUFFICIENT.

Together 1 and 2
x = 4 only

SUFFICIENT.

Answer C.
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Quote:
What is the value of x?

(1) \( \sqrt{x}=x−2\)

(2) \( x ≠ 1\)

(1) \( \sqrt{x}=x−2\) after solved => x=1 or x=4
but \( \sqrt{1}=1−2 = -1\) => nonsense => x=1 was eliminated
=> x=4
=> (1) suff

(2) not suff

=> Choice A
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(1) x can be either 1 or 4. not sufficient.

(2) x can be any number except 1. not sufficient.

Together, x is 4. sufficient.
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We are to determine the value of x.
1. √x=x-2
x=x^2-4x+4
x^2-5x+4=0
(x-1)(x-4)=0
x=1, x=4
Statement 1 alone is insufficient.

2. x≠1
Not sufficient. Infinite possibilities.

1+2
sufficient. x=4.

The answer is C.
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(1) √x=x−2......FROM THIS WE GET X=4,1....Insufficient
(2) x ≠ 1....Clearly insufficient

Combining both we get x=4

OA:C
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C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

\sqrt{x}=x-2

\(x=x^2-4x+4\)
\(0=x^2-5x+4\)
0=(x-1)(x-4)

so must be x=1 or x=4

as Per (2) ≠ 1

SO X = 4
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Both statements are required

Posted from my mobile device
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chetan2u
Quote:
What is the value of x?

(1) \(\sqrt{x} = x-2\)

(2) x ≠ 1

I am pretty sure that when solutions open up, C would win over A and we would learn that how important it is to CHECK the values back in the equation.

Okay so here we are..

(1) \(\sqrt{x} = x-2\)
The first take \(x-2\geq 0\), as \(\sqrt{x}\geq 0\)...So \(x\geq 2\)
square both sides....
\(\sqrt{x} = x-2.........x=(x-2)^2=x^2-4x+4........x^2-5x+4=0......(x-4)(x-1)=0\), so x can be 1 or 4.
But we know that \(x\geq 2\), so x =4
Suff

(2) x ≠ 1
Insuff

A


As x is perfect square we can say x>0 but how root(x)>0. Can you please help.
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chetan2u
Quote:
What is the value of x?

(1) \(\sqrt{x} = x-2\)

(2) x ≠ 1

I am pretty sure that when solutions open up, C would win over A and we would learn that how important it is to CHECK the values back in the equation.

Okay so here we are..

(1) \(\sqrt{x} = x-2\)
The first take \(x-2\geq 0\), as \(\sqrt{x}\geq 0\)...So \(x\geq 2\)
square both sides....
\(\sqrt{x} = x-2.........x=(x-2)^2=x^2-4x+4........x^2-5x+4=0......(x-4)(x-1)=0\), so x can be 1 or 4.
But we know that \(x\geq 2\), so x =4
Suff

(2) x ≠ 1
Insuff

A


As x is perfect square we can say x>0 but how root(x)>0. Can you please help.

Square root is always positive.
\(\sqrt{X}\) is always non negative.
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This cant be a 700-750 level question

Posted from my mobile device
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maitryyadav
This cant be a 700-750 level question

Posted from my mobile device

We do not assign the difficulty level manually. The difficulty level of a question on the site is determined automatically based on various parameters collected from users' attempts, such as the percentage of correct answers and the time taken to answer the question. You can find the difficulty level of a question and its related statistics in the first post.

Hope it clarifies.
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X=?

1)
\(\sqrt{x}=x-2\)

\(0=x-\sqrt{x}-2\)

Let \(y=\sqrt{x}\)

\(y^2-y-2=0\)

\((y-2)(y+1)=0\)


\(y=2\) or \(y=-1\)

Recall that y is the square root of x, which must take a positive value. Therefore, we can discard the solution y=-1. We are left with y=2, which gives us a value of x=4. Sufficient.

2) Clearly ins.

A.
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