If (4 + x^1/2) =(x + 2) , then x could be equal to which of the foll

Question

If  \sqrt{4 + x^{\frac{1}{2}}} =\sqrt{x + 2} , then x could be equal to which of the following?

A. -1
B. 0
C. 1
D. 4
E. cannot be determined.

PareshGmat · Accepted Answer

\sqrt{4 + x^{\frac{1}{2}}} =\sqrt{x + 2}

Squaring both sides

4 + x^{\frac{1}{2}} = x+2

x - \sqrt{x} = 2

Only for x = 4, the above equation would hold true

Answer = D

BrentGMATPrepNow · Answer

As has already been demonstrated, the fastest solution is to plug in each answer choice. This allows us to solve the question in well under a minute. 
That said, we can also solve the equation algebraically.

For equations involving square roots, we must make sure we test for  extraneous roots  (more in the video below).

Given:  \sqrt{4 + x^{\frac{1}{2}}} =\sqrt{x + 2}

Square both sides to get: 4 + x^(1/2) = x + 2
Subtract 4 from both sides: x^(1/2) = x - 2
Square both sides: x = (x - 2)²
Expand: x = x² - 4x + 4
Subtract x from both sides: 0 = x² - 5x + 4
Factor: 0 = (x - 1)(x - 4)
So, EITHER x = 1 OR x = 4

IMPORTANT: Now plug the solutions into the original equation to check for  extraneous roots.

Try x = 1
 Given: \sqrt{4 + x^{\frac{1}{2}}} =\sqrt{x + 2} 
Replace x with 1 to get:  \sqrt{4 + 1^{\frac{1}{2}}} =\sqrt{1 + 2} 
Simplify:  \sqrt{5} =\sqrt{3} 
Doesn't work. 
ELIMINATE x = 1

Try x = 4
 Given: \sqrt{4 + x^{\frac{1}{2}}} =\sqrt{x + 2} 
Replace x with 4 to get:  \sqrt{4 + 4^{\frac{1}{2}}} =\sqrt{4 + 2} 
Simplify:  \sqrt{4 + 2} =\sqrt{4 + 2} 
Simplify:  \sqrt{6} =\sqrt{6} 
WORKS!!

Answer:  D

RELATED VIDEO
 https://www.youtube.com/watch?v=Hr3T9Bi7wkU

manpreetsingh86 · Answer

\sqrt{4 + x^{\frac{1}{2}}} =\sqrt{x + 2}

squaring both sides we have

4 + x^1/2= x+2
=x-x^1/2-2
let x^1/2=k
x= (x^1/2)^2 = k^2

k^2-k-2=0
k^2-2k+k-2=0
k(k-2)+1(k-2)
(k+1)(k-2)=0
k=-1 or k=2
or x^1/2=-1 or x^1/2=2
x^1/2=-1 is not possible. as square root of x will be positive.
thus x^1/2 = 2
squaring both sides we have
x=4

dinesh86 · Answer

No need to solve...just put the options one by one and check. Only 4 make sense . LHS=RHS

Ashishmathew01081987 · Answer

Plug in the choices in place of x

We need LHS = RHS

Only 4 makes the LHS = RHS

So answer is D

2013gmat · Answer

√(4 + x^1/2) =√(x + 2)

squaring both sides..

4 + x^1/2 = x+2
x^1/2 - x-2

SBS again..

x = x^2 +4 - 4x
x=1,4

can not be determined.

But if we plug in the values, x = 4.. clear answer.

Bunuel, please tell me where I am lacking in terms of approach.

Thanks!

Ashishmathew01081987 · Answer

Yes your method is correct...

I had made a mistake once and i forgot to plug in the values of X in the equation.

Here You have to plug in the values of X =1,4 in the equation to check its validity.

If X=1 then LHS is not equal to RHS,  Hence not valid 
If X=4 then LHS = RHS,  Hence Valid

shreyast · Answer

As far as I know,  x^{1/2} , can yield both +ve and -ve roots, unlike  \sqrt{x}  where only positive roots are possible.

If the OA is correct then the question should be modified to replace  x^{1/2}  by  \sqrt{x} .
Or the OA should be changed to E.

Bunuel , could you please help to comment?

EMPOWERgmatRichC · Answer

Hi shreyast,

I think that you're confusing one rule with another.

If you're given X^2 = 16, then there ARE 2 solutions: +4 and -4

If you're given X^(1/2) = √X = 16, then there is just ONE solution: +4

GMAT assassins aren't born, they're made,
Rich

reto · Answer

Hi Paresh 
Could you kindly show me how you would solve the equiton the get the final possible values of X?
Thank you

Bunuel · Answer

x - \sqrt{x} = 2 ;

x - \sqrt{x} -2= 0 ;

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

Solve quadratic for  \sqrt{x} :

\sqrt{x}=-1  (discard since the square root of a number cannot be negative) or  \sqrt{x}=2 .

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If (4 + x^1/2) =(x + 2) , then x could be equal to which of the foll

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If (4 + x^1/2) =(x + 2) , then x could be equal to which of the foll

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