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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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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
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Bunuel wrote:

Tough and Tricky questions: Algebra.

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.

Kudos for a correct solution.

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!!

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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
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Bunuel wrote:

Tough and Tricky questions: Algebra.

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.

Kudos for a correct solution.

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

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
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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
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No need to solve...just put the options one by one and check. Only 4 make sense . LHS=RHS
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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
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Bunuel wrote:

Tough and Tricky questions: Algebra.

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.

Kudos for a correct solution.

Plug in the choices in place of x

We need LHS = RHS

Only 4 makes the LHS = RHS

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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
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√(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!
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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
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2013gmat wrote:
√(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!

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
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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
2013gmat wrote:
√(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!

This is correct. However, after obtaining values of x = 1,4, we require to return back to the "original equation" to check the feasibility

For x = 1 will have LHS NOT equal to RHS

For x = 4, LHS = RHS which holds true
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If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
x=4 makes sense only after ignoring the possibility that sqrt(4) = -2. In the same sense, option C (x=1) also correct if we consider the situation that sqrt(1)= -1 . This is definitely not a 700 level question, but surely a controversial one.
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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
I also used the same way as Paresh. Worked pretty fast. Less than 1 min for sure.
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If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of th [#permalink]
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.

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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
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

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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal [#permalink]
PareshGmat wrote:
$$\sqrt{4 + x^{\frac{1}{2}}} =\sqrt{x + 2}$$

Squaring both sides

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

$$x - \sqrt{x} = 2$$

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

Hi Paresh
Could you kindly show me how you would solve the equiton the get the final possible values of X?
Thank you
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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
manpreetsingh86 wrote:
Bunuel wrote:

Tough and Tricky questions: Algebra.

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.

Kudos for a correct solution.

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

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

Very Very lengthy
Just plug in the values..
it will spare some tim
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If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
Hi Guys,

When we square both sides of the equation, isn't the result in the absolute value form? such as --> l 4 + x^1/2 l = l x + 2 l ? I am confused with this. As (√x)^2 = lxl since we don't know if x is positive or negative. Why this rule is not applied here?

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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
HarveyKlaus wrote:
Hi Guys,

When we square both sides of the equation, isn't the result in the absolute value form? such as --> l 4 + x^1/2 l = l x + 2 l ? I am confused with this. As (√x)^2 = lxl since we don't know if x is positive or negative. Why this rule is not applied here?

Hi
Why are you using the absolute value theory here
Just plug in the options.
and to point out => x has to be positive here as x^1/2 is √x which cant be -ve in the gmat maths

Regards
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Re: If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
How do we know when to try feeding in the values from given options and when to solve the equation to derive the value of x?
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If √(4 + x^1/2) =√(x + 2) , then x could be equal to which of the foll [#permalink]
How do we know when to try feeding in the values from given options and when to solve the equation to derive the value of x?

I would say , you could only decide this by looking at the question.

$$\sqrt{4 + x^1/2}$$ = $$\sqrt{x + 2}$$

First thing you should remember seeing a square root is that the variable under square root can never be negative.so if you know this fact you can straight away get away with the first option.
Now ask yourself whether you want to solve this equation as it has $$\sqrt{x}$$ and x as variable - so you are talking about a quadratic equation.

now look at the answer choices : its easier to substitute and check the values

[B] 0 : 2 =$$\sqrt{2}$$ ; not true
[C] 1 : $$\sqrt{5}$$ = $$\sqrt{3}$$ ; not true
[D] 4 : $$\sqrt{6}$$ = $$\sqrt{6}$$ - bingo! no need to check option E
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