GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 22 Jan 2019, 08:54

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
  • The winners of the GMAT game show

     January 22, 2019

     January 22, 2019

     10:00 PM PST

     11:00 PM PST

    In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.
  • Key Strategies to Master GMAT SC

     January 26, 2019

     January 26, 2019

     07:00 AM PST

     09:00 AM PST

    Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

D01-13

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52386
D01-13  [#permalink]

Show Tags

New post 15 Sep 2014, 23:11
4
27
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

35% (01:30) correct 65% (01:03) wrong based on 339 sessions

HideShow timer Statistics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52386
Re D01-13  [#permalink]

Show Tags

New post 15 Sep 2014, 23:11
6
10
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: \(x^4=x^3+6x^2\);

Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);

Factorize: \(x^2(x-3)(x+2)=0\);

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
_________________

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?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 29 Apr 2014
Posts: 67
Location: India
Concentration: Other, Finance
GMAT 1: 770 Q51 V42
WE: Analyst (Retail Banking)
Re: D01-13  [#permalink]

Show Tags

New post 02 Oct 2014, 06:12
2
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: \(x^4=x^3+6x^2\);

Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);

Factorize: \(x^2(x-3)(x+2)=0\);

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
User avatar
V
Joined: 02 Sep 2009
Posts: 52386
Re: D01-13  [#permalink]

Show Tags

New post 02 Oct 2014, 14:16
5
2
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: \(x^4=x^3+6x^2\);

Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);

Factorize: \(x^2(x-3)(x+2)=0\);

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?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 23 Jan 2013
Posts: 138
Concentration: Technology, Other
Schools: Berkeley Haas
GMAT Date: 01-14-2015
WE: Information Technology (Computer Software)
GMAT ToolKit User
Re: D01-13  [#permalink]

Show Tags

New post 17 Oct 2014, 21:44
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: \(x^4=x^3+6x^2\);

Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);

Factorize: \(x^2(x-3)(x+2)=0\);

[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
User avatar
V
Joined: 02 Sep 2009
Posts: 52386
Re: D01-13  [#permalink]

Show Tags

New post 18 Oct 2014, 01:11
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: \(x^4=x^3+6x^2\);

Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);

Factorize: \(x^2(x-3)(x+2)=0\);

[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?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
User avatar
Joined: 02 Jul 2012
Posts: 187
Location: India
Schools: IIMC (A)
GMAT 1: 720 Q50 V38
GPA: 2.6
WE: Information Technology (Consulting)
Reviews Badge
Re: D01-13  [#permalink]

Show Tags

New post 23 Oct 2014, 03:00
1
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, \(x^4= x^3+6x^2\)
\(=>\) \(x^4-x^3-6x^2=0\)
\(x^2(x^2-x-6)=0\)
\(x=0,3,-2\)

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... :-D

Intern
Intern
avatar
Joined: 26 Jun 2014
Posts: 1
D01-13  [#permalink]

Show Tags

New post 31 Oct 2014, 15:26
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. \(x^2-x-6=0\)?

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
Manager
User avatar
Joined: 02 Jul 2012
Posts: 187
Location: India
Schools: IIMC (A)
GMAT 1: 720 Q50 V38
GPA: 2.6
WE: Information Technology (Consulting)
Reviews Badge
Re: D01-13  [#permalink]

Show Tags

New post 31 Oct 2014, 19:59
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. \(x^2-x-6=0\)?

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... :-D

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52386
D01-13  [#permalink]

Show Tags

New post 01 Nov 2014, 03:51
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. \(x^2-x-6=0\)?

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: algebra-tips-and-hints-175003.html

Hope 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?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 30 Mar 2013
Posts: 109
GMAT ToolKit User
Re: D01-13  [#permalink]

Show Tags

New post 08 Dec 2014, 12:31
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
Intern
avatar
Joined: 10 Dec 2014
Posts: 29
GMAT Date: 12-30-2014
Re: D01-13  [#permalink]

Show Tags

New post 14 Dec 2014, 21:16
We need not even consider -2 and check it.Because the root will never give a negative value
Current Student
avatar
B
Joined: 15 Apr 2015
Posts: 13
Location: India
Schools: EDHEC (A$)
GMAT 1: 620 Q44 V31
GPA: 3.25
Re D01-13  [#permalink]

Show Tags

New post 10 Sep 2015, 06:19
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
User avatar
V
Joined: 02 Sep 2009
Posts: 52386
Re: D01-13  [#permalink]

Show Tags

New post 10 Sep 2015, 07:19
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?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 05 Aug 2015
Posts: 42
Re: D01-13  [#permalink]

Show Tags

New post 02 Nov 2015, 08:36
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
Manager
avatar
B
Joined: 27 Aug 2014
Posts: 74
Re: D01-13  [#permalink]

Show Tags

New post 26 Feb 2016, 06:20
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^2-x-6=0 that should have only two roots. So that should give some hint that all three roots cant be acceptable.
Intern
Intern
avatar
Joined: 07 Dec 2015
Posts: 2
Location: Nepal
Concentration: Finance, Statistics
GPA: 3
Re D01-13  [#permalink]

Show Tags

New post 17 Mar 2016, 07:24
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
User avatar
V
Joined: 02 Sep 2009
Posts: 52386
Re: D01-13  [#permalink]

Show Tags

New post 17 Mar 2016, 09:27
2
1
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?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
B
Joined: 11 Jan 2015
Posts: 34
GMAT ToolKit User
Re: D01-13  [#permalink]

Show Tags

New post 26 Mar 2016, 15:24
Hi Bunuel, would the following be a valid approach to this question or just luck?

1.) Re-arrange 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
User avatar
V
Joined: 02 Sep 2009
Posts: 52386
Re: D01-13  [#permalink]

Show Tags

New post 27 Mar 2016, 01:47
paddy41 wrote:
Hi Bunuel, would the following be a valid approach to this question or just luck?

1.) Re-arrange 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


You should find the actual roots and find out whether all of them are valid. Only 2 of them are valid for the equation at hand. Please refer to the discussion on previous 2 pages.
_________________

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?
Extra-hard Quant Tests with Brilliant Analytics

GMAT Club Bot
Re: D01-13 &nbs [#permalink] 27 Mar 2016, 01:47

Go to page    1   2   3    Next  [ 47 posts ] 

Display posts from previous: Sort by

D01-13

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel



Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.