Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44321

New Algebra Set!!! [#permalink]
Show Tags
18 Mar 2013, 07:56
31
This post received KUDOS
Expert's post
153
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: 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: 26 Feb 2015
Posts: 123

Re: New Algebra Set!!! [#permalink]
Show Tags
10 May 2015, 23:06
1
This post received KUDOS
Bunuel wrote: Please suggest on what category would you like the next set to be. Thank you! Hey Bunuel, I'd really like to see a Fraction/Decimal/Percent type of set. Off topic, but just a heads up, under your "RESOURCES", you got Remainders mentioned twice (7 and 10)



Manager
Joined: 18 Aug 2014
Posts: 128
Location: Hong Kong

Re: New Algebra Set!!! [#permalink]
Show Tags
26 May 2015, 11:21
Bunuel wrote: 6. 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
Rearrange: \((mn)^2 + mn  12=0\).
Factorize for mn: \((mn+4)(mn3)=0\). Thus \(mn=4\) or \(mn=3\).
So, we have that \(m=\frac{4}{n}\) or \(m=\frac{3}{n}\).
Answer: E. Hello, how come m^2n^2 will become (mn)^2 ?



Math Expert
Joined: 02 Sep 2009
Posts: 44321

Re: New Algebra Set!!! [#permalink]
Show Tags
27 May 2015, 03:28



Manager
Joined: 26 Dec 2012
Posts: 148
Location: United States
Concentration: Technology, Social Entrepreneurship
WE: Information Technology (Computer Software)

Re: New Algebra Set!!! [#permalink]
Show Tags
27 May 2015, 11:26
1. Double square both side we will get, x^4=x^3+6x^2; rearranging we get x^2(x^2x6) =0 Solving above equation, x=0,2 &3 Putting all these values in original expression we get, root under (expression); and GMAT does not support negative value in side root. So possible value of x=0 &3 Hence sum of all possible solution =3, Answer is D
9. If x=(√5−√7)2, then the best approximation of x is X= 5+72*root 5*root7 =12 2*root35 approximately =122*6 =1212 =0 Hence answer is A



Manager
Joined: 26 Dec 2012
Posts: 148
Location: United States
Concentration: Technology, Social Entrepreneurship
WE: Information Technology (Computer Software)

Re: New Algebra Set!!! [#permalink]
Show Tags
27 May 2015, 13:48
10. 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) ? Replace n^2 instead of x in f(x) term and (n+12)^2 instead of x in g(x), we will get 2n^21=(n+12)^2 = n^2+144+24n, solving equation we get n^224n145 =0 Resolving above equation we get n= 5 &29 Hence product of all values of n =5*29 = 145; hence answer is A
8. If m is a negative integer and m^3 + 380 = 381m , then what is the value of m? Rearranging above equation we get: m^3381m+380=0 or m(m^2 – 381) +380=0 So any value of m which satisfy “ m(m^2 381) = 380 “ equation will be correct value of x. Only m=20 satisfy the given equation , hence m=20 , answer is B
7. If x^4 = 29x^2  100, then which of the following is NOT a product of three possible values of x? Assume x^2 =n ; solve equation n^229n+100=0 therefore n=5 &25 As n=x^2 = 25 &4 , x=+5,5,+2 & 2. So we can have product of possible values as 25, 50,+50 ,…..etc but x can never be +25 So Answer is B
6. If mn does not equal to zero, and m^2n^2 + mn = 12, then m could be: Assume mn =x; so we will have x^2+x12=0 solving for above equation we get x= 4 & 3 So mn=4 & 3 therefore m= 4/n & 3/n Hence values of m is 1 and 3, answer is E
4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ? As x^2 and y^2 terms are in negative so any value (positive or negative) will give negative and minimize the value of entire term. So making x & y zero will give maximum value of the term hence 39 is the maximum value , answer is A
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: Adding both equation we get 2 a^2 = (m+n) => a= root[ (m+n)/2] Subtracting both equation we get 2 b ^2 = (mn) =>b= root[ (mn)/2] Therefore a*b = root [(m+n)(mn)] /2 = ½ * root (m^2 –n^2) Hence answer is C
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? Substitution and finding out value of a from x=3; 9+3a+15=0 => a= 8 As x^2+ax –b =0 has equal roots, so b^2 4ac =0 therefore a^2+ 4b=0 => a^2= 4b Substitution value of a in above equation. b=  (8*8)/4 = 16 Hence answer is B



Intern
Joined: 27 Apr 2015
Posts: 8

Re: New Algebra Set!!! [#permalink]
Show Tags
27 May 2015, 13:52
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, I understand how you got to the three roots: 0, 3, and 2 but I missed out on the 0 root because I factored the expression like this: x^4 = x^3 + 6x^2 x^4 = x^2(x + 6) x^2 = x + 6 x^2  x  6 = 0 .... leading to only roots 3 and 2 I'm not sure why my way is incorrect since. I don't know if I broke some algebra rule. Could you please clarify my factoring method? Thanks, kindly.



Manager
Joined: 18 Aug 2014
Posts: 128
Location: Hong Kong

Re: New Algebra Set!!! [#permalink]
Show Tags
27 May 2015, 23:13
Bunuel wrote: LaxAvenger wrote: Bunuel wrote: 6. 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
Rearrange: \((mn)^2 + mn  12=0\).
Factorize for mn: \((mn+4)(mn3)=0\). Thus \(mn=4\) or \(mn=3\).
So, we have that \(m=\frac{4}{n}\) or \(m=\frac{3}{n}\).
Answer: E. Hello, how come m^2n^2 will become (mn)^2 ? \((mn)^2=(mn)(mn)=m^2*n^2\). Ohh.. the writing is a little bit misleading. I was thinking of: m^2n^2 and not m^2 * n^2



Math Expert
Joined: 02 Sep 2009
Posts: 44321

New Algebra Set!!! [#permalink]
Show Tags
28 May 2015, 04:18
torontoclub15 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, I understand how you got to the three roots: 0, 3, and 2 but I missed out on the 0 root because I factored the expression like this: x^4 = x^3 + 6x^2 x^4 = x^2(x + 6) x^2 = x + 6 x^2  x  6 = 0 .... leading to only roots 3 and 2 I'm not sure why my way is incorrect since. I don't know if I broke some algebra rule. Could you please clarify my factoring method? Thanks, kindly. If you divide (reduce) x^4 = x^2(x + 6) by x^2, you assume, with no ground for it, that x^2 (x) does not equal to zero thus exclude a possible solution. 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.
_________________
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: 31 Jul 2014
Posts: 143

Re: New Algebra Set!!! [#permalink]
Show Tags
11 Sep 2015, 08:17
Bunuel wrote: Sorry, there was a typo in the stem .
5. 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
Rearrange the given equation: \(x^22x+15=m\).
Given that \(x\) is an integer from 10 and 10, inclusive (21 values) we need to find the probability that \(x^22x+15\) is greater than zero, so the probability that \(x^22x+15>0\).
Factorize: \((x+5)(3x)>0\). This equation holds true for \(5<x<3\).
Since x is an integer then it can take the following 7 values: 4, 3, 2, 1, 0, 1, and 2.
So, the probability is 7/21=1/3.
Answer: B. Hi Bunuel Could you please check below is fine? Thanks can we write ? m>0 > x^22x+15 > 0 x^2+2x15 < 0 (x+5)(x3) < 0 test some big number like 100 > (x+5)(x3) will be positive  +ve 5  ve 3 +ve range of x ==> 5<x<3



Director
Affiliations: GMATQuantum
Joined: 19 Apr 2009
Posts: 607

Re: New Algebra Set!!! [#permalink]
Show Tags
11 Sep 2015, 22:18
Hi anupamadw,
Yes, that works. When you have a quadratic inequality and there are two real roots, the number line can be thought of being divided in to three regions. One to the left of x=5, the region between 5 and 3, and the region to the right of 3. The sign of the expression will change as we go from one region to the next. This means our expression will either follow the path of +, , + or , +,  in the three regions. When you chose x=100 as a sample value, we found that the expression (x+5)(x3) is positive at x=100 and we can conclude that for x>3 it is positive, this means that between 5 and 3 it will be negative, and less 5 it will be positive.
Cheers, Dabral



Manager
Joined: 02 Jun 2015
Posts: 191
Location: Ghana

New Algebra Set!!! [#permalink]
Show Tags
08 Feb 2016, 10:07
Bunuel 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
Rearrange and factor for x^2: \((x^225)(x^24)=0\).
So, we have that \(x=5\), \(x=5\), \(x=2\), or \(x=2\).
\(50=5*(5)*2\); \(50=5*(5)*(2)\).
Only 25 is NOT a product of three possible values of x
Answer: B. Hi Bunuel, Can you kindly help me to understand why you included the negative values of X in calculating the product of possible values of X. Can you explain that visavis question 1's solution.
_________________
Kindly press kudos if you find my post helpful



Intern
Joined: 06 Mar 2016
Posts: 2

Re: New Algebra Set!!! [#permalink]
Show Tags
10 Dec 2016, 14:08
[quote="Bunuel"] 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. Bunuel can u please explain why it isn't C? I mean why can't x be a negative number, when in fact positive or negative x^4 both can be the given expression right? So 3 minus 2 gives us 1



Math Expert
Joined: 02 Sep 2009
Posts: 44321

Re: New Algebra Set!!! [#permalink]
Show Tags
10 Dec 2016, 15:33
dinobull 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. Bunuel can u please explain why it isn't C? I mean why can't x be a negative number, when in fact positive or negative x^4 both can be the given expression right? So 3 minus 2 gives us 1 Please read the previous discussion. Your doubt was addressed several times before.
_________________
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: 27 Jan 2017
Posts: 1

Re: New Algebra Set!!! [#permalink]
Show Tags
28 Jan 2017, 04:23
WoundedTiger 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
Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get
3^2 +3a+15=0> Solving for a we get a=8.
Putting the value if a in Eqn 1 we get x^2 8xb=0 We know for Eqn ax^2 +bx+c=0 Sum of roots is given by b/a Product of roots is c/a
Therefore from Eqn1 (after subsitution of value of a) we get Sum of roots = 8 and roots are equal ie. 4 and 4 Product of roots will be 16. Therefore Option B Hello How do I know that I have to plug in 3 for X if it says that the root of the equation is 3. I dont understand. Can someone explain? Thank you very much



Math Expert
Joined: 02 Sep 2009
Posts: 44321

Re: New Algebra Set!!! [#permalink]
Show Tags
28 Jan 2017, 07:52
sonnenblume92 wrote: WoundedTiger 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
Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get
3^2 +3a+15=0> Solving for a we get a=8.
Putting the value if a in Eqn 1 we get x^2 8xb=0 We know for Eqn ax^2 +bx+c=0 Sum of roots is given by b/a Product of roots is c/a
Therefore from Eqn1 (after subsitution of value of a) we get Sum of roots = 8 and roots are equal ie. 4 and 4 Product of roots will be 16. Therefore Option B Hello How do I know that I have to plug in 3 for X if it says that the root of the equation is 3. I dont understand. Can someone explain? Thank you very much 3 is a root of x^2 + ax + 15 = 0 means that when you plug 3 there the equation must hold true.
_________________
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: 10 Jun 2016
Posts: 52

New Algebra Set!!! [#permalink]
Show Tags
29 Jan 2017, 13:20
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. Don't understand why to to maximize the value of \(3(x2)^22(y+3)^2\) must be Zero ??
_________________
Thank You Very Much, CoolKl Success is the Journey from Knowing to Doing
A Kudo is a gesture, to express the effort helped. Thanks for your Kudos.



Math Expert
Joined: 02 Sep 2009
Posts: 44321

Re: New Algebra Set!!! [#permalink]
Show Tags
31 Jan 2017, 10:52
coolkl wrote: 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. Don't understand why to to maximize the value of \(3(x2)^22(y+3)^2\) must be Zero ?? The max value of 3(x2)^2 is 0 and max value of 2(y+3)^2 is 0 too. You see since the square of a number is always non negative, then 2*(y+3)^2 = negative*nonnegative = nonposiitve (so 2(y+3)^2 is negative or zero).
_________________
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: 17 Apr 2016
Posts: 102

Re: New Algebra Set!!! [#permalink]
Show Tags
31 Jan 2017, 20:08
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 Bunnel, As per the explanation the maximum value of the expresssion is zero because the expressions in the brackets is squared (implying it will always be positive ) and the expression is multiplied by a negative number. But what if these were the expressions: \(3(x2)^2\) and \(2(y+3)^2\) How can we find the maximum values of such expressions? Thanks. Adi



Math Expert
Joined: 02 Sep 2009
Posts: 44321

Re: New Algebra Set!!! [#permalink]
Show Tags
01 Feb 2017, 08:09
adityapareshshah wrote: 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 Bunnel, As per the explanation the maximum value of the expresssion is zero because the expressions in the brackets is squared (implying it will always be positive ) and the expression is multiplied by a negative number. But what if these were the expressions: \(3(x2)^2\) and \(2(y+3)^2\) How can we find the maximum values of such expressions? Thanks. Adi In this case there won't be a maximum value.
_________________
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: 29 Jul 2016
Posts: 8

Re: New Algebra Set!!! [#permalink]
Show Tags
13 Mar 2017, 09:17
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: \(3^2+3a+15=0\). 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. Hi Bunuel, how come I cannot/should not solve this problem by treating them as if they are two equations and multiple unknowns? As in, x^2 + ax + 15 = 0 ; x^2 +ax b = 0. Subtract the two equations and you get 15+b=0 so b= 15. Thanks!




Re: New Algebra Set!!!
[#permalink]
13 Mar 2017, 09:17



Go to page
Previous
1 2 3 4 5 6 7 8 9 10 11
Next
[ 210 posts ]



