Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47981

Question Stats:
36% (01:39) correct 64% (01:06) wrong based on 295 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 47981

Re D0113
[#permalink]
Show Tags
16 Sep 2014, 00:11



Manager
Joined: 29 Apr 2014
Posts: 67
Location: India
Concentration: Other, Finance
WE: Analyst (Retail Banking)

Re: D0113
[#permalink]
Show Tags
02 Oct 2014, 07:12
Bunuel wrote: Official Solution:
If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. \(2\) B. \(0\) C. \(1\) D. \(3\) E. \(5\)
Take the given expression to the 4th power: \(x^4=x^3+6x^2\); Rearrange and factor out x^2: \(x^2(x^2x6)=0\); Factorize: \(x^2(x3)(x+2)=0\); So, the roots are \(x=0\), \(x=3\) and \(x=2\). But \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\). The sum of all possible solutions for x is 0+3=3.
Answer: D Why can 4th root of an expression not be negative? \({2}^4 = 16\) and \(2^4 = 16\) So, shouldn't \(\sqrt[4]{16}\) be \(2\) or \(2\)?
_________________
Psst...If that helped press the thumb on the left



Math Expert
Joined: 02 Sep 2009
Posts: 47981

Re: D0113
[#permalink]
Show Tags
02 Oct 2014, 15:16
eshan429 wrote: Bunuel wrote: Official Solution:
If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. \(2\) B. \(0\) C. \(1\) D. \(3\) E. \(5\)
Take the given expression to the 4th power: \(x^4=x^3+6x^2\); Rearrange and factor out x^2: \(x^2(x^2x6)=0\); Factorize: \(x^2(x3)(x+2)=0\); So, the roots are \(x=0\), \(x=3\) and \(x=2\). But \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\). The sum of all possible solutions for x is 0+3=3.
Answer: D Why can 4th root of an expression not be negative? \({2}^4 = 16\) and \(2^4 = 16\) So, shouldn't \(\sqrt[4]{16}\) be \(2\) or \(2\)? First of all \({2}^4 = 16\), it would be 16 if it were \({(2)}^4\). Next, 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, \(\sqrt{25}=5\), NOT +5 or 5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5. Even roots have only a positive value on the GMAT.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 23 Jan 2013
Posts: 163
Concentration: Technology, Other
GMAT Date: 01142015
WE: Information Technology (Computer Software)

Re: D0113
[#permalink]
Show Tags
17 Oct 2014, 22:44
Bunuel wrote: Official Solution:
If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. \(2\) B. \(0\) C. \(1\) D. \(3\) E. \(5\)
Take the given expression to the 4th power: \(x^4=x^3+6x^2\); Rearrange and factor out x^2: \(x^2(x^2x6)=0\); Factorize: \(x^2(x3)(x+2)=0\); [highlight]So, the roots are \(x=0\), \(x=3\) and \(x=2\). But \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\).[/highlight] The sum of all possible solutions for x is 0+3=3.
Answer: D Hi Bunnel , Could you elaborate more on the above statement ? If i substitute X = 0 , 2 , 3 Equation turns out to be 0 in all cases which it statisfies hence i feel all answers are correct . Maybe my thinking is flawed . Thanks and Regards , Shelrod007



Math Expert
Joined: 02 Sep 2009
Posts: 47981

Re: D0113
[#permalink]
Show Tags
18 Oct 2014, 02:11
shelrod007 wrote: Bunuel wrote: Official Solution:
If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. \(2\) B. \(0\) C. \(1\) D. \(3\) E. \(5\)
Take the given expression to the 4th power: \(x^4=x^3+6x^2\); Rearrange and factor out x^2: \(x^2(x^2x6)=0\); Factorize: \(x^2(x3)(x+2)=0\); [highlight]So, the roots are \(x=0\), \(x=3\) and \(x=2\). But \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\).[/highlight] The sum of all possible solutions for x is 0+3=3.
Answer: D Hi Bunnel , Could you elaborate more on the above statement ? If i substitute X = 0 , 2 , 3 Equation turns out to be 0 in all cases which it statisfies hence i feel all answers are correct . Maybe my thinking is flawed . Thanks and Regards , Shelrod007 Yes, your are not right. Have you read the whole thread? How does 2 satisfy \(x=\sqrt[4]{x^3+6x^2}\)?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 02 Jul 2012
Posts: 202
Location: India
GPA: 2.6
WE: Information Technology (Consulting)

Re: D0113
[#permalink]
Show Tags
23 Oct 2014, 04:00
Bunuel wrote: If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. \(2\) B. \(0\) C. \(1\) D. \(3\) E. \(5\) I got this question wrong in the test. So, solving it here for a better understanding. Basically this question tests one concept i.e. \(\sqrt{x^2}\) is x and not x. Solving the equation, \(x^4= x^3+6x^2\) \(=>\) \(x^4x^36x^2=0\) \(x^2(x^2x6)=0\) \(x=0,3,2\) As mentioned earlier, \(\sqrt{x^2}\) is x and not x, the value 2 is discarded. Thus the two values are 3 and 0 SO, the sum = 3
_________________
Give KUDOS if the post helps you...



Intern
Joined: 26 Jun 2014
Posts: 1

would it be wrong to have \(x^4 = x^3+6x^2\) (as previously stated) then factor out \(x^2\) on both sides for\((x^2)(x^2) = (x^2)(x+6)\) then divide both sides by \(x^2\) leaving \(x^2=x+6\), ie. \(x^2x6=0\)?
if this is acceptable does that still leave 0 as a solution with 2 and 3? I'm more curious if this would cause problems on other similar problems.



Manager
Joined: 02 Jul 2012
Posts: 202
Location: India
GPA: 2.6
WE: Information Technology (Consulting)

Re: D0113
[#permalink]
Show Tags
31 Oct 2014, 20:59
SQUEE wrote: would it be wrong to have \(x^4 = x^3+6x^2\) (as previously stated) then factor out \(x^2\) on both sides for\((x^2)(x^2) = (x^2)(x+6)\) then divide both sides by \(x^2\) leaving \(x^2=x+6\), ie. \(x^2x6=0\)?
if this is acceptable does that still leave 0 as a solution with 2 and 3? I'm more curious if this would cause problems on other similar problems. Dear SQUEE, If you would divide both sides by x^2 this would even contain the possibility of dividing by 0. As we know that we can not divide by 0 since division by 0 is not defined. So do dividing by x^2 would be wrong. Hope that helps you.
_________________
Give KUDOS if the post helps you...



Math Expert
Joined: 02 Sep 2009
Posts: 47981

SQUEE wrote: would it be wrong to have \(x^4 = x^3+6x^2\) (as previously stated) then factor out \(x^2\) on both sides for\((x^2)(x^2) = (x^2)(x+6)\) then divide both sides by \(x^2\) leaving \(x^2=x+6\), ie. \(x^2x6=0\)?
if this is acceptable does that still leave 0 as a solution with 2 and 3? I'm more curious if this would cause problems on other similar problems. Also, if you divide (reduce) x^4 = x^2(x + 6) by x^2 you assume, with no ground for it, that x does not equal to zero thus exclude a possible solution (notice that x=0 satisfies the equation). Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.Check more tips on Algebra here: algebratipsandhints175003.htmlHope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 30 Mar 2013
Posts: 113

Re: D0113
[#permalink]
Show Tags
08 Dec 2014, 13:31
if we take x=2, the solving the equation, LHS would have 2, and the RHS would have a positive 2. Therefore, they are not equal and 2 doesn't satisfy the equation. Plugging the numbers back in help.



Intern
Joined: 10 Dec 2014
Posts: 30
GMAT Date: 12302014

Re: D0113
[#permalink]
Show Tags
14 Dec 2014, 22:16
We need not even consider 2 and check it.Because the root will never give a negative value



Intern
Joined: 15 Apr 2015
Posts: 13
Location: India
GPA: 3.25

Re D0113
[#permalink]
Show Tags
10 Sep 2015, 07:19
I don't agree with the explanation. i don't agree with the explanation of why you have chosen to not include 2 in the sum. Also after taking the 4th power of the equation why can't we use the general formula for n powered equation where sum of roots =b/a. i got 1 using that formula



Math Expert
Joined: 02 Sep 2009
Posts: 47981

Re: D0113
[#permalink]
Show Tags
10 Sep 2015, 08:19
mrigoel wrote: I don't agree with the explanation. i don't agree with the explanation of why you have chosen to not include 2 in the sum. Also after taking the 4th power of the equation why can't we use the general formula for n powered equation where sum of roots =b/a. i got 1 using that formula When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive root. That is: \(\sqrt{9} = 3\), NOT +3 or 3; \(\sqrt[4]{16} = 2\), NOT +2 or 2; Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and 3. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 05 Aug 2015
Posts: 46

Re: D0113
[#permalink]
Show Tags
02 Nov 2015, 09:36
I got x=0, 3, 2 but failed to check to see if all the solutions are valid. Do I always need to check if the roots are valid or just on certain types of problems? If so, which types / under what conditions? Just want to better understand and save time on the test. Thanks so much Gmatclub!
_________________
Working towards 25 Kudos for the Gmatclub Exams  help meee I'm poooor



Manager
Joined: 27 Aug 2014
Posts: 81

Re: D0113
[#permalink]
Show Tags
26 Feb 2016, 07:20
Bunuel wrote: If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. \(2\) B. \(0\) C. \(1\) D. \(3\) E. \(5\) Hi Also my little bit. When reducing, we hit a quadratic equation x^2x6=0 that should have only two roots. So that should give some hint that all three roots cant be acceptable.



Intern
Joined: 07 Dec 2015
Posts: 2
Location: Nepal
Concentration: Finance, Statistics
GPA: 3

Re D0113
[#permalink]
Show Tags
17 Mar 2016, 08:24
I think this the explanation isn't clear enough, please elaborate. i dont understand why 2 is not a solution . for instance squrt of 4 has 2 soln i.e 2 and 2. please explain?



Math Expert
Joined: 02 Sep 2009
Posts: 47981

Re: D0113
[#permalink]
Show Tags
17 Mar 2016, 10:27
sabinadhikari99 wrote: I think this the explanation isn't clear enough, please elaborate. i dont understand why 2 is not a solution . for instance squrt of 4 has 2 soln i.e 2 and 2. please explain? That's not true. 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, \(\sqrt{25}=5\), NOT +5 or 5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5. Even roots have only a positive value on the GMAT.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 11 Jan 2015
Posts: 35

Re: D0113
[#permalink]
Show Tags
26 Mar 2016, 16:24
Hi Bunuel, would the following be a valid approach to this question or just luck?
1.) Rearrange and factor out \(x^2: x^2(x^2−x−6)=0\) > So one solution is 0 > \(x^2\) 2.) Use discriminant for the left equation \((x^2−x−6)=0\): \(b^2  4ac\) > 1(4*1*(6)) > 25 > Since 25 is positive the left equation has two solutions 3.) 1.) + 2.) = 3 Solutions



Math Expert
Joined: 02 Sep 2009
Posts: 47981

Re: D0113
[#permalink]
Show Tags
27 Mar 2016, 02:47







Go to page
1 2 3
Next
[ 46 posts ]



