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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, 07:34
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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, 08:56
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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, 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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If x^2 − 2 < 0, which of the following specifies all the possible [#permalink]
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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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16 Oct 2015, 12:02
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y=x^22=(xsqrt2)*(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, 12:13
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/986Cheers, 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, 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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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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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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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 positivex < √2 When x is negativex < √2 x > √2 Combining these two inequalities, we have: √2 < x < √2 Answer: C
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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, 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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03 Feb 2017, 03:21
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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05 Feb 2018, 12:30
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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