What is the value of x? (1) x^(1/2) = x - 2 (2) x ≠ 1.

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!

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

(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)

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.

(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

(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.

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.

(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

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

Both statements are required

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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.

This cant be a 700-750 level question

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Hope it clarifies.

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.