Oct 14 08:00 PM PDT  11:00 PM PDT Join a 4day FREE online boot camp to kick off your GMAT preparation and get you into your dream bschool in R2.**Limited for the first 99 registrants. Register today! Oct 15 12:00 PM PDT  01:00 PM PDT Join this live GMAT class with GMAT Ninja to learn to conquer your fears of long, kooky GMAT questions. Oct 16 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 29 Jul 2016
Posts: 5

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!



Math Expert
Joined: 02 Sep 2009
Posts: 58313

Re: New Algebra Set!!!
[#permalink]
Show Tags
13 Mar 2017, 09:36
flexman wrote: 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! Because those are not simultaneous equations, those two are not given to be a system of equations.
_________________



Intern
Joined: 10 Jun 2016
Posts: 45

Re: New Algebra Set!!!
[#permalink]
Show Tags
18 Mar 2017, 11:38
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. Hello Bunnel, I did the same way as above except when x^22x+15 > 0, so took ve sign common as x^2+2x15 < 0 (x+5)(x3) < 0 x+5 < 0 or x 3 < 0 x < 5 or x < 3. So for me possible values are 6,78,9,10. Means 5/21. Any idea where am i wrong ?
_________________
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: 58313

Re: New Algebra Set!!!
[#permalink]
Show Tags
19 Mar 2017, 04:16
coolkl wrote: 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. Hello Bunnel, I did the same way as above except when x^22x+15 > 0, so took ve sign common as x^2+2x15 < 0 (x+5)(x3) < 0 x+5 < 0 or x 3 < 0 x < 5 or x < 3. So for me possible values are 6,78,9,10. Means 5/21. Any idea where am i wrong ? When you say that 6,78,9,10 satisfy the inequality, did you test any of these values? Ho did you solve (x+5)(x3) < 0 to get x < 5 or x < 3? What does x < 5 OR x < 3 even mean? It does not make sense... (x+5)(x3) < 0 holds true for 5 < x < 3. Check the links below for more: Inequality tipsInequalities Made Easy!Solving Quadratic Inequalities  Graphic ApproachWavy Line Method Application  Complex Algebraic InequalitiesDS Inequalities Problems PS Inequalities Problems
_________________



Intern
Joined: 07 Sep 2016
Posts: 29

Re: New Algebra Set!!!
[#permalink]
Show Tags
14 Jun 2017, 21:09
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. You have been clearing doubts , one more need your help. WHy X cannot take negative value? What I understand that If X=3 then (3)^4=(3)^4 .



Math Expert
Joined: 02 Sep 2009
Posts: 58313

Re: New Algebra Set!!!
[#permalink]
Show Tags
14 Jun 2017, 22:19
gmat4varun 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. You have been clearing doubts , one more need your help. WHy X cannot take negative value? What I understand that If X=3 then (3)^4=(3)^4 . Please read the whole thread. \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)). 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 a positive value on the GMAT.In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5.
_________________



Intern
Joined: 18 Jan 2017
Posts: 43
Location: India
Concentration: General Management, Entrepreneurship
GPA: 4

Re: New Algebra Set!!!
[#permalink]
Show Tags
12 Dec 2017, 08:53
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. How did you get this directly?



Math Expert
Joined: 02 Sep 2009
Posts: 58313

Re: New Algebra Set!!!
[#permalink]
Show Tags
12 Dec 2017, 08:57
srishti201996 wrote: 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. How did you get this directly? Factoring QuadraticsSolving Quadratic Equations For more: 7. Algebra Other topics: Ultimate GMAT Quantitative MegathreadHope it helps.
_________________



Intern
Joined: 22 Oct 2017
Posts: 5

Re: New Algebra Set!!!
[#permalink]
Show Tags
06 Feb 2018, 10:21
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, hope you can help: In the description it says "has equal roots, and one of the roots of the equation is ... 3". I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" > therefore i would have never thought about setting the determinant equal to zero. Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero? Thank you very much! Best regards



Math Expert
Joined: 02 Sep 2009
Posts: 58313

Re: New Algebra Set!!!
[#permalink]
Show Tags
06 Feb 2018, 10:24
andr3 wrote: 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, hope you can help: In the description it says "has equal roots, and one of the roots of the equation is ... 3". I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" > therefore i would have never thought about setting the determinant equal to zero. Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero? Thank you very much! Best regards 1. x^2 + ax  b = 0 has equal roots 2. One of the roots of the equation x^2 + ax + 15 = 0 is 3. Those are two different equations.
_________________



Intern
Joined: 22 Oct 2017
Posts: 5

Re: New Algebra Set!!!
[#permalink]
Show Tags
06 Feb 2018, 10:42
Bunuel wrote: andr3 wrote: 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, hope you can help: In the description it says "has equal roots, and one of the roots of the equation is ... 3". I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" > therefore i would have never thought about setting the determinant equal to zero. Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero? Thank you very much! Best regards 1. x^2 + ax  b = 0 has equal roots 2. One of the roots of the equation x^2 + ax + 15 = 0 is 3. Those are two different equations. ok, than it is that language of "has equal roots", which I don't understand. could you clarify what "equal roots" means? is "equal roots" the definition of having only one solution? thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 58313

Re: New Algebra Set!!!
[#permalink]
Show Tags
06 Feb 2018, 10:44
andr3 wrote: ok, than it is that language of "has equal roots", which I don't understand. could you clarify what "equal roots" means? is "equal roots" the definition of having only one solution?
thanks!
Yes, equal roots means that there are two roots which are equal to each other, so one distinct root.
_________________



Intern
Joined: 09 Feb 2018
Posts: 5

Re: New Algebra Set!!!
[#permalink]
Show Tags
22 Feb 2018, 22:47
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. can someone help me how the max value of the highlighted text is defined ?



Math Expert
Joined: 02 Sep 2009
Posts: 58313

Re: New Algebra Set!!!
[#permalink]
Show Tags
22 Feb 2018, 22:53
SaravanaPrabu090492 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. can someone help me how the max value of the highlighted text is defined ? \(3(x2)^2= (3)*(square \ of \ a \ number)= (negative)*(nonnegative) = (nonpositive)\). So, this term is negative or 0. The same for \(2(y+3)^2\).
_________________



Intern
Joined: 09 Feb 2018
Posts: 5

Re: New Algebra Set!!!
[#permalink]
Show Tags
22 Feb 2018, 23:42
Bunuel wrote: 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) ?
A. 145 B. 24 C. 24 D. 145 E. None of the above
\(f(x) = 2x  1\), hence \(f(n^2)=2n^21\). \(g(x) = x^2\), hence \(g(n+12)=(n+12)^2=n^2+24n+144\).
Since given that \(f(n^2)=g(n+12)\), then \(2n^21=n^2+24n+144\). Rearranging gives \(n^224n145=0\).
Next, Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):
\(x_1+x_2=\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).
Thus according to the above \(n_1*n_2=145\).
Answer: A. Having a doubt with the above highlighted text. for the first, shouldn't the whole term [2x1] be squared like the below one ? seeking solutions. thanks in advance.



Math Expert
Joined: 02 Sep 2009
Posts: 58313

Re: New Algebra Set!!!
[#permalink]
Show Tags
23 Feb 2018, 00:15
SaravanaPrabu090492 wrote: Bunuel wrote: 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) ?
A. 145 B. 24 C. 24 D. 145 E. None of the above
\(f(x) = 2x  1\), hence \(f(n^2)=2n^21\). \(g(x) = x^2\), hence \(g(n+12)=(n+12)^2=n^2+24n+144\).
Since given that \(f(n^2)=g(n+12)\), then \(2n^21=n^2+24n+144\). Rearranging gives \(n^224n145=0\).
Next, Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):
\(x_1+x_2=\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).
Thus according to the above \(n_1*n_2=145\).
Answer: A. Having a doubt with the above highlighted text. for the first, shouldn't the whole term [2x1] be squared like the below one ? seeking solutions. thanks in advance. No. \(f(x) = 2x  1\) means that we have a function which tells us what to do to get the value of the function: take a value for which you want to find the output of the function, multiply it by 2 and subtract 1. So, to get \(f(n^2)\) put n^2 instead of x: \(f(n^2)=2n^21\). Similarly for \(g(x) = x^2\). To get \(g(n+12)\) put n + 12 instead of x: \(g(n+12)=(n+12)^2=n^2+24n+144\).
_________________



Manager
Status: The journey is always more beautiful than the destination
Affiliations: Computer Science
Joined: 24 Apr 2017
Posts: 50
Location: India
Concentration: Statistics, Strategy
GPA: 3.14

Re: New Algebra Set!!!
[#permalink]
Show Tags
05 Apr 2018, 09:40
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. \(3x^2 + 12x 2y^2  12y  39\) >\(x(123x)  y(2y+12)39\) now x(123x) will be highest when x is 1. y(2y+12) will be higest when y = 1. so the answer becomes 9+1039 = 19.. where am I going wrong?
_________________
Sky is the limit. 800 is the limit.



Manager
Status: The journey is always more beautiful than the destination
Affiliations: Computer Science
Joined: 24 Apr 2017
Posts: 50
Location: India
Concentration: Statistics, Strategy
GPA: 3.14

Re: New Algebra Set!!!
[#permalink]
Show Tags
05 Apr 2018, 09:51
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. Quote: Factorize: \((x+5)(3x)>0\). This equation holds true for \(5<x<3\) unable to understand this line. Can someone please explain? many thanks
_________________
Sky is the limit. 800 is the limit.



Math Expert
Joined: 02 Sep 2009
Posts: 58313

Re: New Algebra Set!!!
[#permalink]
Show Tags
05 Apr 2018, 13:27
Nixondutta wrote: 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. Quote: Factorize: \((x+5)(3x)>0\). This equation holds true for \(5<x<3\) unable to understand this line. Can someone please explain? many thanks Check this: Solving Quadratic Inequalities: Graphic ApproachFro more: check Inequalities For more check Ultimate GMAT Quantitative Megathread Hope it helps.
_________________



Intern
Joined: 27 Jan 2016
Posts: 22
Location: India
Concentration: International Business, Operations
WE: Design (Energy and Utilities)

Re: New Algebra Set!!!
[#permalink]
Show Tags
24 Nov 2018, 13:27
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. Can anybody help me with this solution. Why the maximum value of \(3(x2)^2\) and \(2(y+3)^2\) is zero?




Re: New Algebra Set!!!
[#permalink]
24 Nov 2018, 13:27



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



