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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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16 Oct 2015, 06:34
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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.
[Reveal] Spoiler: OA

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

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16 Oct 2015, 07:56
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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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If x^2 − 2 < 0, which of the following specifies all the possible [#permalink]

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16 Oct 2015, 08:19
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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 $$-\sqrt{2} < x <\sqrt{2}$$

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

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16 Oct 2015, 09:40
x^$${2}$$ -2 < 0
=> x^$${2}$$<2

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

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

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16 Oct 2015, 11:02
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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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16 Oct 2015, 11:13
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Expert's post
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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If x^2 − 2 < 0, which of the following specifies all the possible [#permalink]

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17 Oct 2015, 06:23
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..
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Re: If x^2 − 2 < 0, which of the following specifies all the possible [#permalink]

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18 Oct 2015, 09:04
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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..

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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02 May 2016, 02:56
$$x^2$$ - 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 <$$\sqrt{2}$$
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Re: If x^2 − 2 < 0, which of the following specifies all the possible [#permalink]

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02 May 2016, 06:17
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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

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

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02 Feb 2017, 12:58
\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..

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?
Math Expert
Joined: 02 Sep 2009
Posts: 43867
Re: If x^2 − 2 < 0, which of the following specifies all the possible [#permalink]

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03 Feb 2017, 02:21
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Expert's post
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..

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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05 Feb 2018, 11:30
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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05 Feb 2018, 11:34
Rocket7 wrote:
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

If you take the square root from x^2 < 2 you get $$|x| < \sqrt{2}$$, which in turn gives $$-\sqrt{2} < x <\sqrt{2}$$.
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Re: If x^2 − 2 < 0, which of the following specifies all the possible   [#permalink] 05 Feb 2018, 11:34
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