Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 39622

New Algebra Set!!! [#permalink]
Show Tags
18 Mar 2013, 07:56
32
This post received KUDOS
Expert's post
118
This post was BOOKMARKED
The next set of medium/hard PS algebra questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.1. 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 Solution: newalgebraset14934960.html#p12009482. The equation x^2 + ax  b = 0 has equal roots, and one of the roots of the equation x^2 + ax + 15 = 0 is 3. What is the value of b?A. 64 B. 16 C. 15 D. 1/16 E. 1/64 Solution: newalgebraset14934960.html#p12009503. If a and b are positive numbers, such that a^2 + b^2 = m and a^2  b^2 = n, then ab in terms of m and n equals to:A. \(\frac{\sqrt{mn}}{2}\) B. \(\frac{\sqrt{mn}}{2}\) C. \(\frac{\sqrt{m^2n^2}}{2}\) D. \(\frac{\sqrt{n^2m^2}}{2}\) E. \(\frac{\sqrt{m^2+n^2}}{2}\) Solution: newalgebraset14934960.html#p12009564. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?A. 39 B. 9 C. 0 D. 9 E. 39 Solution: newalgebraset14934960.html#p12009625. If x^2 + 2x 15 = m, where x is an integer from 10 and 10, inclusive, what is the probability that m is greater than zero?A. 2/7 B. 1/3 C. 7/20 D. 2/5 E. 3/7 Solution: newalgebraset14934960.html#p12009706. If mn does not equal to zero, and m^2n^2 + mn = 12, then m could be:I. 4/n II. 2/n III. 3/n A. I only B. II only C. III only D. I and II only E. I and III only Solution: newalgebraset14934960.html#p12009737. If x^4 = 29x^2  100, then which of the following is NOT a product of three possible values of x? I. 50 II. 25 III. 50 A. I only B. II only C. III only D. I and II only E. I and III only Solution: newalgebraset14934960.html#p12009758. If m is a negative integer and m^3 + 380 = 381m , then what is the value of m?A. 21 B. 20 C. 19 D. 1 E. None of the above Solution: newalgebraset14934960.html#p12009809. If \(x=(\sqrt{5}\sqrt{7})^2\), then the best approximation of x is:A. 0 B. 1 C. 2 D. 3 E. 4 Solution: newalgebraset14934960.html#p120098210. If f(x) = 2x  1 and g(x) = x^2, then what is the product of all values of n for which f(n^2)=g(n+12) ?A. 145 B. 24 C. 24 D. 145 E. None of the above Solution: newalgebraset14934980.html#p1200987Kudos points for each correct solution!!!
_________________
New to the Math Forum? Please read this: 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



Current Student
Joined: 04 Mar 2013
Posts: 69
Location: India
Concentration: Strategy, Operations
GPA: 3.66
WE: Operations (Manufacturing)

Re: New Algebra Set!!! [#permalink]
Show Tags
12 Apr 2013, 05:25
Narenn wrote: nave81 wrote: Bunuel wrote: SOLUTIONs:
1. 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. Hi Bunuel, Could you please explain why 2 is not a solution? Is it because x is a nonnegative number? Because if I plug in the value of 2 to the equation \(x^4=x^3+6x^2\) I get \(2=\sqrt[4]{16}\) which can be true isn't it, as \(\sqrt[4]{16}=2\) Also could you explain why even root of an expression cannot be negative when we have an expression which is a square, because we know that \(\sqrt{{x^2}}=x\)? Thank you Note the Important Difference. On the GMAT if X^2 = 4 then x = +/ 2 or x = 2 However if x= sq.root(4) then x has to be positive i.e. 2 and it can not take negative value. when we plug 2 as the value of x in the equation we would get 2 = 4th root of 16 > 2 = 2 This is because fourth root of 16 is 2 and not 2 The rule is even root of a number can not be negative on the GMAT Regards, Abhijit. Hey Abhijit, This is a very interesting point that you have made here. This statement that 2 is the fourth root of 16 is admissible under normal circumstances but is not in a GMAT question. Can you mention a source which would list out such peculiar rules for GMAT quants? Thanks in advance Anshuman
_________________
When you feel like giving up, remember why you held on for so long in the first place.



MBA Section Director
Status: Back to work...
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 4395
Location: India
City: Pune
GPA: 3.4
WE: Business Development (Manufacturing)

Re: New Algebra Set!!! [#permalink]
Show Tags
13 Apr 2013, 09:01
Hi Anshuman, For this you need to go thru some strategy books of GMAT Quant. In My Opinion, two important sources are, 1) Manhattan GMAT Quant strategy guides 4th or 5th Ed 2) GMAT Club Math Book  This is free and you can download it from this website free of cost (Check BB's profile) Furthermore you can check my signature. There i have attached my articles prepared on some peculiar topics. You may find them useful. Thanks, Abhijit
_________________
Every Wednesday: Meet MBA Experts in Chat Room and Ask Your MostPressing MBA Admission Questions to them in a Live Chat.
Must Read Forum Topics Before You Kick Off Your MBA Application
New GMAT Club Decision Tracker  Real Time Decision Updates



Senior Manager
Joined: 07 Sep 2010
Posts: 327

Re: New Algebra Set!!! [#permalink]
Show Tags
22 May 2013, 22:19
Bunuel wrote: [ 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. Hi Bunuel, Request you to please let me know where I'm making a mistake. Since \(x^3+6*x^2 >= 0\) Then, \(x^2(x+6)>=0\) i.e \((x0)^2(x(6))>=0\) This implies that\( x>=6\) Hence, x=2 is a valid root, and sum of all roots should be\(x=3+(2) = 1\) Please let me know where I am going wrong. Thanks
_________________
+1 Kudos me, Help me unlocking GMAT Club Tests



Math Expert
Joined: 02 Sep 2009
Posts: 39622

Re: New Algebra Set!!! [#permalink]
Show Tags
23 May 2013, 02:00
imhimanshu wrote: Bunuel wrote: [ 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. Hi Bunuel, Request you to please let me know where I'm making a mistake. Since \(x^3+6*x^2 >= 0\) Then, \(x^2(x+6)>=0\) i.e \((x0)^2(x(6))>=0\) This implies that\( x>=6\) Hence, x=2 is a valid root, and sum of all roots should be\(x=3+(2) = 1\) Please let me know where I am going wrong. Thanks Plug x=2 into \(x=\sqrt[4]{x^3+6x^2}\). Does the equation hold true?
_________________
New to the Math Forum? Please read this: 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: 13 May 2013
Posts: 12

Re: New Algebra Set!!! [#permalink]
Show Tags
23 May 2013, 05:03
2. The equation x^2 + ax  b = 0 has equal roots, and one of the roots of the equation x^2 + ax + 15 = 0 is 3. What is the value of b?
A. 64 B. 16 C. 15 D. 1/16 E. 1/64
Solution:
x^2 + ax + 15 = 0 equation has one root as 3. only possible values of getting product of the roots 15 , having one root 3 is 5*3. x^25x3x+15=0 take common values out => x(x5)3(x5) =0 => x=5 and x=3(what was given) , from this we get a= 8.
x^2 8x  b =0 => if roots are equal then b^24ac=0, 64+4b=0 from this b=16
Answer B



Math Expert
Joined: 02 Sep 2009
Posts: 39622

Re: New Algebra Set!!! [#permalink]
Show Tags
23 May 2013, 05:42
SathyaNIT wrote: 7. If x^4 = 29x^2  100, then which of the following is NOT a product of three possible values of x? I. 50 II. 25 III. 50 A. I only B. II only C. III only D. I and II only E. I and III only Solution:By using below substitution values we could reduce the powers of x. Here values (1,0,29,0,100)  taken from equation  x^4 = 29x^2  100=> (1)x^4+(0)x^329x^2+(0)x+100=0 from above explanation we get equation x^225=0 => give x=+5,5 from above calculation we get values of x =+2,2. multiplying any of the three roots we have possibility of getting product values, 50,+50. Ans:E The correct answer is B, not E: newalgebraset14934960.html#p1200975Notice that we are asked "which of the following is NOT a product of three possible values of x". Hope it helps.
_________________
New to the Math Forum? Please read this: 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
Status: Working hard to score better on GMAT
Joined: 02 Oct 2012
Posts: 90
Location: Nepal
Concentration: Finance, Entrepreneurship
GPA: 3.83
WE: Accounting (Consulting)

Re: New Algebra Set!!! [#permalink]
Show Tags
25 May 2013, 23:27
Bunuel wrote: 3. If a and b are positive numbers, such that a^2 + b^2 = m and a^2  b^2 = n, then ab in terms of m and n equals to:
A. \(\frac{\sqrt{mn}}{2}\) B. \(\frac{\sqrt{mn}}{2}\) C. \(\frac{\sqrt{m^2n^2}}{2}\) D. \(\frac{\sqrt{n^2m^2}}{2}\) E. \(\frac{\sqrt{m^2+n^2}}{2}\)
Sum the two equations: \(2a^2=m+n\); Subtract the two equations: \(2b^2=mn\);
Multiply: \(4a^2b^2=m^2n^2\);
Solve for \(ab\): \(ab=\frac{\sqrt{m^2n^2}}{2}\)
Answer: C. Another method is picking numbers: let a=2 b=3 m=a^2+b^2 ie 2*2+3*3= 13 Similarly, n= a^2b^2 ie 2*23*3= 5 Now we have to find a*b ie 2*3=6 plug the values of m=13 and n=5 Option C gives 6 which we are looking for.... Thus Answer is C..
_________________
Do not forget to hit the Kudos button on your left if you find my post helpful.



Intern
Joined: 26 Jul 2012
Posts: 2

Re: New Algebra Set!!! [#permalink]
Show Tags
04 Jun 2013, 23:45
In question no.6) how is m^2n^2=(mn)^2 ? Someone help me.



Math Expert
Joined: 02 Sep 2009
Posts: 39622

Re: New Algebra Set!!! [#permalink]
Show Tags
05 Jun 2013, 00:40



Intern
Joined: 26 Jul 2012
Posts: 2

Re: New Algebra Set!!! [#permalink]
Show Tags
05 Jun 2013, 07:08
thank you Bunuel,,, I read the question differently ( m raised to 2n raised to 2)



Intern
Joined: 11 Nov 2012
Posts: 12

Re: New Algebra Set!!! [#permalink]
Show Tags
23 Jun 2013, 00:35
Hi,
For the first question, what do you mean by "But X cannot be negative as it equals to the even (4th) root of some expression"?



Math Expert
Joined: 02 Sep 2009
Posts: 39622

Re: New Algebra Set!!! [#permalink]
Show Tags
23 Jun 2013, 01:37
aquax wrote: Hi,
For the first question, what do you mean by "But X cannot be negative as it equals to the even (4th) root of some expression"? \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt[even]{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\). 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. Even roots have only nonnegative value on the GMAT.In contrast, the equation \(x^2=25\) has TWO solutions, \(\sqrt{25}=+5\) and \(\sqrt{25}=5\). Hope it helps.
_________________
New to the Math Forum? Please read this: 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



Math Expert
Joined: 02 Sep 2009
Posts: 39622

Re: New Algebra Set!!! [#permalink]
Show Tags
09 Jul 2013, 00:40



Director
Joined: 03 Aug 2012
Posts: 893
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29 GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)

Re: New Algebra Set!!! [#permalink]
Show Tags
21 Jul 2013, 22:05
Bunuel wrote: 4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?
A. 39 B. 9 C. 0 D. 9 E. 39
\(3x^2 + 12x 2y^2  12y  39=3x^2 + 12x122y^2  12y189=3(x^24x+4)2(y^2+6y+9)9=\) \(=3(x2)^22(y+3)^29\).
So, we need to maximize the value of \(3(x2)^22(y+3)^29\).
Since, the maximum value of \(3(x2)^2\) and \(2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+09=9\).
Answer: B. Hi Bunuel, Just as a matter of fact can you please explain that how you reached to factorization. In other words how a candidate will decide how to split 39 between (X,Y). Your reply is appreciated !! Rgds, TGC !!
_________________
Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________



Manager
Joined: 26 Sep 2013
Posts: 220
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41

Re: New Algebra Set!!! [#permalink]
Show Tags
10 Oct 2013, 17:57
Bunuel wrote: 2. The equation x^2 + ax  b = 0 has equal roots, and one of the roots of the equation x^2 + ax + 15 = 0 is 3. What is the value of b?
A. 64 B. 16 C. 15 D. 1/16 E. 1/64
Since one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3, then substituting we'll get: \(3^3+3a+15=0\). Solving for \(a\) gives \(a=8\).
Substitute \(a=8\) in the first equation: \(x^28xb=0\).
Now, we know that it has equal roots thus its discriminant must equal to zero: \(d=(8)^2+4b=0\). Solving for \(b\) gives \(b=16\).
Answer: B. Can you explain what a root of an equation is, as well as a discriminant, I didn't get anything done for this one



Math Expert
Joined: 02 Sep 2009
Posts: 39622

Re: New Algebra Set!!! [#permalink]
Show Tags
11 Oct 2013, 02:49



Math Expert
Joined: 02 Sep 2009
Posts: 39622

Re: New Algebra Set!!! [#permalink]
Show Tags
24 Oct 2013, 01:01
NeetiGupta wrote: Bunuel wrote: Bunuel wrote: 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.
Hi Bunuel, Request you to please let me know where I'm making a mistake.
Since \(x^3+6*x^2 >= 0\) Then, \(x^2(x+6)>=0\)
i.e \((x0)^2(x(6))>=0\) This implies that\(x>=6\) Hence, x=2 is a valid root, and sum of all roots should be\(x=3+(2) = 1\) Please let me know where I am going wrong.
Thanks Plug x=2 into \(x=\sqrt[4]{x^3+6x^2}\). Does the equation hold true? Hi Brunel if say x=16. x^1/2 has two values +4 & 4. So why x^1/4 cannot have +2 & 2 as as values ? Please explain First of all it's 4th root not 2nd root. Next, \(\sqrt[4]{16}=2\), not +2 or 2. 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 nonnegative value on the GMAT.Hope it helps.
_________________
New to the Math Forum? Please read this: 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



Senior Manager
Joined: 20 Dec 2013
Posts: 267
Location: India

Re: New Algebra Set!!! [#permalink]
Show Tags
05 Apr 2014, 03:42
Q1. Option D. Raising both LHS and RHS to fourth power,and making RHS=0 we get \(x^4x^36x^2=0\) Taking \(x^2\) common and factorizing, \(x^2(x+2)(x3)=0\) Three possible values of \(x\) are \(2,3\) and \(0\) But \(x\) can't be \(ve\). So sum of possible values=\(3\)



Senior Manager
Joined: 20 Dec 2013
Posts: 267
Location: India

Re: New Algebra Set!!! [#permalink]
Show Tags
05 Apr 2014, 03:47
Q2 Option B. From second equation,calculate value of \(a\) by putting \(x=3\) \(a=8\) For roots to be equal,\(D=0\) \(a^2+4b=0\) \(b=16\)
Last edited by AKG1593 on 05 Apr 2014, 03:54, edited 1 time in total.



Senior Manager
Joined: 20 Dec 2013
Posts: 267
Location: India

Re: New Algebra Set!!! [#permalink]
Show Tags
05 Apr 2014, 03:54
Q3. Option C. \(a^2+b^2=m\) \(a^2b^2=n\) \(=>a^2=m+n/2\) Or \(a=\sqrt{(m+n)/2}\)
Similarly \(b=\sqrt{(mn)/2}\) Multiplying \(a*b\)\(=\sqrt{(m^2n^2)/4}\)




Re: New Algebra Set!!!
[#permalink]
05 Apr 2014, 03:54



Go to page
Previous
1 2 3 4 5 6 7 8 9 10
Next
[ 194 posts ]




