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If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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20 Jan 2017, 20:01
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If x<0, what is the value of \(\sqrt[4]{(x3)^4} + \sqrt{xx}\) A. 3 B. x+3 C. 32x D. 2x3 E. 2
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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20 Jan 2017, 20:33
Whenever we have absolute value question and the variable is given as negative (like x<0 in this case) then the easiest way to solve the problem is to take x=k (where k is positive), this will make opening absolute values easy [(x3)^4]^1/4 + (xx)^1/2 = [((k)3)^4]^1/4 + ((k)k)^1/2 = [((k+3))^4]^1/4 + ((k)k)^1/2 = [(k+3)^4]^1/4 + ((k)(k))^1/2 [as k is positive so k will be equal to k] = (k+3) + (k^2)^1/2 = k+3 + k = 2k + 3 [x = k => k = x] = 2(x) + 3 = 32x So, Answer is C Hope it helps! siddharthsinha123 wrote: If x<0
What is the value of [(x3)^4]^1/4 + (xx)^1/2
A. 3 B. x+3 C. 32x D. 2x3 E. 2
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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20 Jan 2017, 20:36
siddharthsinha123 wrote: If x<0
What is the value of [(x3)^4]^1/4 + (xx)^1/2
A. 3 B. x+3 C. 32x D. 2x3 E. 2 Hi Best way is to substitute a value for X.. Let x be 2.. So [(23)^4]^1/4+((2)*2)^1/2= ((5)^4)^1/4+4^1/2=5+2=7.. Substitute in choices . A.3. ......NO B. x+3. .....2+3=1...NO C. 32x. .....3(2*2)=3+4=7.. YES D. 2x3. .....2*23=7......NO E. 2. .....NO C
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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22 Jan 2017, 14:51
siddharthsinha123 wrote: If x<0, what is the value of \(\sqrt[4]{(x3)^4} + \sqrt{xx}\)
A. 3 B. x+3 C. 32x D. 2x3 E. 2 Formatted the question.
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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24 Jan 2017, 21:36
chetan2u wrote: siddharthsinha123 wrote: If x<0
What is the value of [(x3)^4]^1/4 + (xx)^1/2
A. 3 B. x+3 C. 32x D. 2x3 E. 2 Hi Best way is to substitute a value for X.. Let x be 2.. So [(23)^4]^1/4+((2)*2)^1/2= ((5)^4)^1/4+4^1/2=5+2=7.. Substitute in choices . A.3. ......NO B. x+3. .....2+3=1...NO C. 32x. .....3(2*2)=3+4=7.. YES D. 2x3. .....2*23=7......NO E. 2. .....NO C Hi, Let me understand that why [(23)^4]^1/4 = 5. I think that [(5)^4]^1/4 should be 5



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If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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05 May 2017, 11:09
abhishekmayank wrote: chetan2u wrote: siddharthsinha123 wrote: If x<0
What is the value of [(x3)^4]^1/4 + (xx)^1/2
A. 3 B. x+3 C. 32x D. 2x3 E. 2 Hi Best way is to substitute a value for X.. Let x be 2.. So [(23)^4]^1/4+((2)*2)^1/2= ((5)^4)^1/4+4^1/2=5+2=7.. Substitute in choices . A.3. ......NO B. x+3. .....2+3=1...NO C. 32x. .....3(2*2)=3+4=7.. YES D. 2x3. .....2*23=7......NO E. 2. .....NO C Hi, Let me understand that why [(23)^4]^1/4 = 5. I think that [(5)^4]^1/4 should be 5 There are 2 rules Here. 1. General Rule of Thumb  Every Number, Positive or Negative, raised to an Even Power will give a Positive Number.Here is the Mathematics of same: \((10)^2 = (10)*(10) = (1)*(1)*(10)*(10) = (1) * (100) = 100\) 2. Rule followed in GMAT If \(x^2 = 16\) x = 4 or +4 But \(\sqrt{16} = +4\)
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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06 May 2017, 07:11
umg wrote: abhishekmayank wrote: chetan2u wrote: [quote="siddharthsinha123"]If x<0
What is the value of [(x3)^4]^1/4 + (xx)^1/2
A. 3 B. x+3 C. 32x D. 2x3 E. 2 Hi Best way is to substitute a value for X.. Let x be 2.. So [(23)^4]^1/4+((2)*2)^1/2= ((5)^4)^1/4+4^1/2=5+2=7.. Substitute in choices . A.3. ......NO B. x+3. .....2+3=1...NO C. 32x. .....3(2*2)=3+4=7.. YES D. 2x3. .....2*23=7......NO E. 2. .....NO C Hi, Let me understand that why [(23)^4]^1/4 = 5. I think that [(5)^4]^1/4 should be 5 There are 2 rules Here. 1. General Rule of Thumb  Every Number, Positive or Negative, raised to an Even Power will give a Positive Number.Here is the Mathematics of same: \((10)^2 = (10)*(10) = (1)*(1)*(10)*(10) = (1) * (100) = 100\) 2. Rule followed in GMAT If \(x^2 = 16\) x = 4 or +4 But \(\sqrt{16} = +4\)[/quote] But still as it is 5^4*1/4 Simplifying it will give 5^1 = 5 I am so confused Sent from my SMN9200 using GMAT Club Forum mobile app



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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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06 May 2017, 08:20
khushbumodi wrote: umg wrote: abhishekmayank wrote: Hi,
Let me understand that why [(23)^4]^1/4 = 5. I think that [(5)^4]^1/4 should be 5
There are 2 rules Here. 1. General Rule of Thumb  Every Number, Positive or Negative, raised to an Even Power will give a Positive Number.Here is the Mathematics of same: \((10)^2 = (10)*(10) = (1)*(1)*(10)*(10) = (1) * (100) = 100\) 2. Rule followed in GMAT If \(x^2 = 16\) x = 4 or +4 But \(\sqrt{16} = +4\) But still as it is 5^4*1/4 Simplifying it will give 5^1 = 5 I am so confused Hmm.. That is a valid point. I did not think of that. What I did was I solved the inside bracket first, applied the power 4 and took the 4th root. I don't know why your method is wrong though mine IS the right way to solve because otherwise we would have 2 different solutions of an operation. I believe that one of the lesser known fallacies of Mathematics is at work here. For example, Here is One of those.. \(\sqrt{x} = \sqrt{(x)*(1)*(1)} = \sqrt{x}*\sqrt{(1)}*\sqrt{(1)} =  \sqrt{x}\) but unless x=0, the above result cannot be true; however since we solved for a general term x, this must be universally true. The fallacy lies in an operation that we did.. We cannot split \(\sqrt{(1)*(1)}\) into \(\sqrt{(1)}*\sqrt{(1)}\) P.S. If you don't know, \((\sqrt{(1)})^2 = (1)\).
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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06 May 2017, 09:43
siddharthsinha123 wrote: If x<0, what is the value of \(\sqrt[4]{(x3)^4} + \sqrt{xx}\)
A. 3 B. x+3 C. 32x D. 2x3 E. 2 Two things to remember: (1) \(\sqrt{x^2}=x\) (2) When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\). Back to the question:According to (1) \(\sqrt[4]{(x3)^4}=x3\). Now, since it's given that x is negative, then x  3 will also be negative, thus according to (2) \(x3 = (x3)=3x\). Next, again since x is negative, then \(x=x\). Thus, \(\sqrt{xx}=\sqrt{(x)(x)}=\sqrt{x^2}=x=x\). So, finally we have \(\sqrt[4]{(x3)^4} + \sqrt{xx}=3xx=32x\). Answer: C.
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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09 May 2017, 11:10
chetan2u wrote: siddharthsinha123 wrote: If x<0
What is the value of [(x3)^4]^1/4 + (xx)^1/2
A. 3 B. x+3 C. 32x D. 2x3 E. 2 Hi Best way is to substitute a value for X.. Let x be 2.. So [(23)^4]^1/4+((2)*2)^1/2= ((5)^4)^1/4+4^1/2=5+2=7.. Substitute in choices . A.3. ......NO B. x+3. .....2+3=1...NO C. 32x. .....3(2*2)=3+4=7.. YES D. 2x3. .....2*23=7......NO E. 2. .....NO C I am confused. How can you substitute x for a negative number, when x<0? wouldn't x have to be positive?



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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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23 Jun 2017, 20:46
Bunuel wrote: siddharthsinha123 wrote: If x<0, what is the value of \(\sqrt[4]{(x3)^4} + \sqrt{xx}\)
A. 3 B. x+3 C. 32x D. 2x3 E. 2 Two things to remember: (1) \(\sqrt{x^2}=x\) (2) When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\). Back to the question:According to (1) \(\sqrt[4]{(x3)^4}=x3\). Now, since it's given that x is negative, then x  3 will also be negative, thus according to (2) \(x3 = (x3)=3x\). Next, again since x is negative, then \(x=x\). Thus, \(\sqrt{xx}=\sqrt{(x)(x)}=\sqrt{x^2}=x=x\). So, finally we have \(\sqrt[4]{(x3)^4} + \sqrt{xx}=3xx=32x\). Answer: C. Hello Bunuel When it's written x, don't we always take the positive value no matter what is inside. Why are we even considering the negative value of x when absolute value of a number is positive. Please explain.



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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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23 Jun 2017, 21:10
Bunuel, abhimahna, BrushMyQuantAbsolute value rule is really creating confusion here. I read that absolute value of an integer/expression is always positive no matter what the sign is inside x. Then can't we just write here: (x3)+ sq. root of (x*x) =x+3+ sq. root of ((x)^2) = x+3+((x)^2)^1/2 =x+3+1*x = x+3+x =3 When I am removing  , I am just considering positive value of whatever is inside  . Please help.
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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23 Jun 2017, 22:10
Hi Shiv, You did everything right apart from opening the second part of the equation sq. root of (x*x) Since x is negative so x = x (since x is negative so x will be positive) sq. root of (x*x) = sq. root of (x*x) = sq. root of (x^2) Now whatever comes out of square root is always positive So sq. root of (x^2) = x Again, x is negative so x = x (since x is negative so x will be positive) So final answer is x + 3  x = 32x Hope it helps! Shiv2016 wrote: Bunuel, abhimahna, BrushMyQuantAbsolute value rule is really creating confusion here. I read that absolute value of an integer/expression is always positive no matter what the sign is inside x. Then can't we just write here: (x3)+ sq. root of (x*x) =x+3+ sq. root of ((x)^2) = x+3+((x)^2)^1/2 =x+3+1*x = x+3+x =3 When I am removing  , I am just considering positive value of whatever is inside  . Please help.
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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23 Jun 2017, 22:32
BrushMyQuant wrote: Hi Shiv, You did everything right apart from opening the second part of the equation sq. root of (x*x) Since x is negative so x = x (since x is negative so x will be positive) sq. root of (x*x) = sq. root of (x*x) = sq. root of (x^2) Now whatever comes out of square root is always positive So sq. root of (x^2) = x Again, x is negative so x = x (since x is negative so x will be positive) So final answer is x + 3  x = 32x Hope it helps! Shiv2016 wrote: Bunuel, abhimahna, BrushMyQuantAbsolute value rule is really creating confusion here. I read that absolute value of an integer/expression is always positive no matter what the sign is inside x. Then can't we just write here: (x3)+ sq. root of (x*x) =x+3+ sq. root of ((x)^2) = x+3+((x)^2)^1/2 =x+3+1*x = x+3+x =3 When I am removing  , I am just considering positive value of whatever is inside  . Please help. Thank you so much for your reply. May I ask a very basic question here? We sat that absolute value of any number is positive. So shouldn't we write only positive value after removing the bars. i.e. x= x It's a very basic question but because of this only I am making mistakes in absolute value questions.



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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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23 Jun 2017, 23:08
x = x if x is positive or zero = x if x is negative or zero Example 3 = 3 = (3) (so x = x if x is negative) Sometimes, it becomes complex to solve questions when x is given as negative. So, to make things simple I suggest to take x = k (where k is positive) and substitute x=k in all the equations. Check my first reply in this thread to understand it better. Hope it helps! Shiv2016 wrote: BrushMyQuant wrote: Hi Shiv, You did everything right apart from opening the second part of the equation sq. root of (x*x) Since x is negative so x = x (since x is negative so x will be positive) sq. root of (x*x) = sq. root of (x*x) = sq. root of (x^2) Now whatever comes out of square root is always positive So sq. root of (x^2) = x Again, x is negative so x = x (since x is negative so x will be positive) So final answer is x + 3  x = 32x Hope it helps! Shiv2016 wrote: Bunuel, abhimahna, BrushMyQuantAbsolute value rule is really creating confusion here. I read that absolute value of an integer/expression is always positive no matter what the sign is inside x. Then can't we just write here: (x3)+ sq. root of (x*x) =x+3+ sq. root of ((x)^2) = x+3+((x)^2)^1/2 =x+3+1*x = x+3+x =3 When I am removing  , I am just considering positive value of whatever is inside  . Please help. Thank you so much for your reply. May I ask a very basic question here? We sat that absolute value of any number is positive. So shouldn't we write only positive value after removing the bars. i.e. x= x It's a very basic question but because of this only I am making mistakes in absolute value questions.
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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23 Jun 2017, 23:23
So what I understood is: Absolute value of x is positive x when x is >=0. Absolute value of x is negative x when x<0. So we can't just say that absolute value of every number is positive no matter what is inside. The sign inside  will matter when  are removed. x when x is positive= x x when x is negative= 1*x= +x
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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23 Jun 2017, 23:28
Shiv2016 wrote: So what I understood is:
Absolute value of x is positive x when x is >=0. Absolute value of x is negative x when x<0.
So we can't just say that absolute value of every number is positive no matter what is inside. The sign inside  will matter when  are removed.
x when x is positive= x x when x is negative= 1*x= +x Your understanding is wrong. The post here: https://gmatclub.com/forum/ifx0what ... l#p1847895 explains it. Absolute value of a number CANNOT be negative! x is ALWAYS more than or equal to zero. some positive value = that positive value some negative value = (that negative value), which still will be positive because negative = positive. 0 = 0. Hope it's clear.
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Re: If x<0 What is the value of [(x3)^4]^1/4 + (xx)^1/2 A. 3 B. x+3
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18 Jul 2017, 03:23
BrushMyQuant wrote: Whenever we have absolute value question and the variable is given as negative (like x<0 in this case) then the easiest way to solve the problem is to take x=k (where k is positive), this will make opening absolute values easy [(x3)^4]^1/4 + (xx)^1/2 = [((k)3)^4]^1/4 + ((k)k)^1/2 = [((k+3))^4]^1/4 + ((k)k)^1/2 = [(k+3)^4]^1/4 + ((k)(k))^1/2 [as k is positive so k will be equal to k] = (k+3) + (k^2)^1/2 = k+3 + k = 2k + 3 [x = k => k = x] = 2(x) + 3 = 32x So, Answer is C Hope it helps! siddharthsinha123 wrote: If x<0
What is the value of [(x3)^4]^1/4 + (xx)^1/2
A. 3 B. x+3 C. 32x D. 2x3 E. 2 Can't we just cross off power 4 and 1/4? x=k [(k3)^4]^1/4 + \sqrt{(k)*k} = k3+ k = 3 I have tried this question so many times but still not really able to understand the minor things. I don't understand where I am going wrong: in cancelling the powers or opening the .



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