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If x^2 − 2 < 0, which of the following specifies all the possible

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If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 16 Oct 2015, 07:34
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A
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C
D
E

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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 02 May 2016, 07:17
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Bunuel wrote:
If x^2 − 2 < 0, which of the following specifies all the possible values of x?

(A) 0 < x < 2
(B) 0 < x <
(C) \(-\sqrt{2} < x <\sqrt{2}\)
(D) −2 < x < 0
(E) −2 < x < 2


Kudos for a correct solution.


Solution:

We must determine all of the values of x based on the inequality x^2 – 2 < 0. To determine all the possible values of x, let’s simplify the inequality.

x^2 < 2

√x^2 < √2

|x| < √2

Note that when we take the square root of x^2, the value of x itself can be either negative or positive, and thus we express this as |x|, the absolute value of x.

Remember also that when we are solving an absolute value equation or inequality, we consider two cases: when the quantity inside the absolute value is positive and then when the quantity inside the absolute value is negative.

Let's now solve for when x is positive and then for when x is negative.

When x is positive

x < √2

When x is negative

-x < √2

x > -√2

Combining these two inequalities, we have:

-√2 < x < √2

Answer: C
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 16 Oct 2015, 09:19
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Bunuel wrote:
If x^2 − 2 < 0, which of the following specifies all the possible values of x?

(A) 0 < x < 2
(B) 0 < x <
(C) \(-\sqrt{2} < x <\sqrt{2}\)
(D) −2 < x < 0
(E) −2 < x < 2



x^2 − 2 < 0
=> x^2 < 2
=> mode of x < \(\sqrt{2}\)

therefore \(-\sqrt{2} < x <\sqrt{2}\)

Answer option C
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 16 Oct 2015, 08:56
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If x^2 − 2 < 0, which of the following specifies all the possible values of x?

we can see that x^2 is less than 2. it can be the case that sqrt(2) must be less than 2.
since the square of any negative integer is a positive number, we should take into consideration that x cannot be less than -sqrt(2), but not greater than sqrt(2)
the only answer here is C
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 16 Oct 2015, 10:40
x^\({2}\) -2 < 0
=> x^\({2}\)<2

Therefore , - \(\sqrt{(2)}\)< x <\(\sqrt{(2)}\)

Answer C
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 16 Oct 2015, 12:02
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y=x^2-2=(x-sqrt2)*(x+sqrt2) when x in in between these two values (-sqrt2,sqrt2) ,y will be negative.
So Option C
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 16 Oct 2015, 12:13
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Bunuel wrote:
If x^2 − 2 < 0, which of the following specifies all the possible values of x?

(A) 0 < x < 2
(B) 0 < x <
(C) \(-\sqrt{2} < x <\sqrt{2}\)
(D) −2 < x < 0
(E) −2 < x < 2


If anyone is interested, we have a free video that explains how to solve quadratic inequalities such as this - http://www.gmatprepnow.com/module/gmat- ... /video/986

Cheers,
Brent
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 17 Oct 2015, 07:23
mvictor wrote:
If x^2 − 2 < 0, which of the following specifies all the possible values of x?

we can see that x^2 is less than 2. it can be the case that sqrt(2) must be less than 2.
since the square of any negative integer is a positive number, we should take into consideration that x cannot be less than -sqrt(2), but not greater than sqrt(2)
the only answer here is C


Hi..
According to me X^2 < 2 can be written as Mod (X) < 2, which in tern yields the range of x as -2 < x < 2... Could you tell me where iam going wrong..
Thanks in advance.. :) :)
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 18 Oct 2015, 10:04
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goldfinchmonster wrote:
mvictor wrote:
If x^2 − 2 < 0, which of the following specifies all the possible values of x?

we can see that x^2 is less than 2. it can be the case that sqrt(2) must be less than 2.
since the square of any negative integer is a positive number, we should take into consideration that x cannot be less than -sqrt(2), but not greater than sqrt(2)
the only answer here is C


Hi..
According to me X^2 < 2 can be written as Mod (X) < 2, which in tern yields the range of x as -2 < x < 2... Could you tell me where iam going wrong..
Thanks in advance.. :) :)


If you take the square root from x^2 < 2 you get \(|x| < \sqrt{2}\), which in turn gives \(-\sqrt{2} < x <\sqrt{2}\).

Hope it's clear.
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 02 May 2016, 03:56
\(x^2\) - 2 < 0
\(x^2\)< 2
As we know roots can be positive as well as as negative
Solution for \(x^2\)< 2
- \(\sqrt{2}\) < x <\(\sqrt{2}\)
correct answer - C
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 02 Feb 2017, 13:58
\sqrt{}
Bunuel wrote:
goldfinchmonster wrote:
mvictor wrote:
If x^2 − 2 < 0, which of the following specifies all the possible values of x?

we can see that x^2 is less than 2. it can be the case that sqrt(2) must be less than 2.
since the square of any negative integer is a positive number, we should take into consideration that x cannot be less than -sqrt(2), but not greater than sqrt(2)
the only answer here is C


Hi..
According to me X^2 < 2 can be written as Mod (X) < 2, which in tern yields the range of x as -2 < x < 2... Could you tell me where iam going wrong..
Thanks in advance.. :) :)


If you take the square root from x^2 < 2 you get \(|x| < \sqrt{2}\), which in turn gives \(-\sqrt{2} < x <\sqrt{2}\).

Hope it's clear.



Hi Bunuel

I'm struggling a little with this one.

Indeed I assumed that when GMAT gives the symbol of a square root, we should only consider the positive number of x. In other words if GMAT asks x^16 therefore X = 4 or x = -4 BUT if GMAT asks √16 therefore we should only consider 4 and NOT -4.
Coming back to the question, I assumed that x^2 < 2 and therefore √x<√2 and therefore x must be strictly positive and be between 0 and √2

Even though I kind of understand your reasoning, where am I doing wrong here?
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 03 Feb 2017, 03:21
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YanisBoubenider wrote:
\sqrt{}
Bunuel wrote:
goldfinchmonster wrote:

Hi..
According to me X^2 < 2 can be written as Mod (X) < 2, which in tern yields the range of x as -2 < x < 2... Could you tell me where iam going wrong..
Thanks in advance.. :) :)


If you take the square root from x^2 < 2 you get \(|x| < \sqrt{2}\), which in turn gives \(-\sqrt{2} < x <\sqrt{2}\).

Hope it's clear.



Hi Bunuel

I'm struggling a little with this one.

Indeed I assumed that when GMAT gives the symbol of a square root, we should only consider the positive number of x. In other words if GMAT asks x^16 therefore X = 4 or x = -4 BUT if GMAT asks √16 therefore we should only consider 4 and NOT -4.
Coming back to the question, I assumed that x^2 < 2 and therefore √x<√2 and therefore x must be strictly positive and be between 0 and √2

Even though I kind of understand your reasoning, where am I doing wrong here?


When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{16}=4\), NOT +4 or -4. Even roots have only a positive value on the GMAT.

In contrast, the equation \(x^2=16\) has TWO solutions, +4 and -4.

Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

Now, about the problem itself: MUST KNOW: \(\sqrt{x^2}=|x|\):

The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

For example if x = -5, then \(\sqrt{x^2}=\sqrt{25}=5=|-5|=|x|\)
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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New post 05 Feb 2018, 12:30
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hello

Can anyone explain why E is wrong? I understand why C is right and I understand the way modulus is functioning in this Q. so when we have x^2<2. cant we just say that x must be -2<x<2
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Re: If x^2 − 2 < 0, which of the following specifies all the possible  [#permalink]

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Re: If x^2 − 2 < 0, which of the following specifies all the possible   [#permalink] 19 Feb 2019, 18:45
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