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New Algebra Set!!!
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Updated on: 18 Feb 2019, 05:31
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: https://gmatclub.com/forum/newalgebra ... l#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: https://gmatclub.com/forum/newalgebra ... l#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: https://gmatclub.com/forum/newalgebra ... l#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: https://gmatclub.com/forum/newalgebra ... l#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: https://gmatclub.com/forum/newalgebra ... l#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: https://gmatclub.com/forum/newalgebra ... l#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: https://gmatclub.com/forum/newalgebra ... l#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: https://gmatclub.com/forum/newalgebra ... l#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: https://gmatclub.com/forum/newalgebra ... l#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: https://gmatclub.com/forum/newalgebra ... l#p1200987Kudos points for each correct solution!!!
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Originally posted by Bunuel on 18 Mar 2013, 07:56.
Last edited by Bunuel on 18 Feb 2019, 05:31, edited 1 time in total.
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Re: New Algebra Set!!!
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22 Mar 2013, 05:01
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.
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Re: New Algebra Set!!!
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18 Mar 2013, 09:03
8. m^3 + 380 = 381m m^3 + 380 = 380m + m m^3 m = 380m  380 m(m1)(m+1) = 380 (m1) m(m+1) = 380 ( 20 * 19; m is negative ) m = 20.
Answer is B




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Re: New Algebra Set!!!
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22 Mar 2013, 06:30
8. 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 Given \(m^3 + 380 = 380m+m\). Rearrange: \(m^3m= 380m380\). \(m(m+1)(m1)=380(m1)\). Since m is a negative integer, then \(m1\neq{0}\) and we can safely reduce by \(m1\) to get \(m(m+1)=380\). So, we have that 380 is the product of two consecutive negative integers: \(380=20*(19)\), hence \(m=20\). Answer: B.
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Re: New Algebra Set!!!
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22 Mar 2013, 05:16
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.
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Re: New Algebra Set!!!
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22 Mar 2013, 06:44
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.
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Re: New Algebra Set!!!
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22 Mar 2013, 06:08
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.
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Re: New Algebra Set!!!
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22 Mar 2013, 05:10
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.
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Re: New Algebra Set!!!
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22 Mar 2013, 05:47
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.
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Re: New Algebra Set!!!
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18 Mar 2013, 08:22
3. a^2 + b^ 2 = m a^2  b^2 = n Solving both the equations( adding them, and then subtracting them ): 2a^2 = m + n 2b^2 = m  n. a = ((m+n)/2)^(1/2) b = ((mn)/2)^(1/2)
ab = ((m^2  n^2)^(1/2))/2
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Re: New Algebra Set!!!
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18 Mar 2013, 09:42
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 \(m^2n^2 + mn = 12\) \(mn(mn+1) = 12\) mn = 3 \(3(3+1)=12\) OR mn = 4 \(4(4+1)=12\) \(m= 3/n\) OR \(m=4/n\) E
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22 Mar 2013, 06:20
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. 50A. 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.
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Re: New Algebra Set!!!
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18 Mar 2013, 08:11
GyanOne wrote: 1. x^4 = x^3 + 6x^2 => x^2 (x^2  x  6) = 0 Sum of roots of x^2  x  6 = 0 is 1 and the only other solution is x=0 => Sum of all possible solutions = 1.
Option C Some questions are tricky!!!
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Re: New Algebra Set!!!
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18 Mar 2013, 09:33
1)\(x^4=x^3+6x^2\) \(x^4x^36x^2=0\) \(x^2(x^2x6)=0\) \(x^2=0 (1) x = 0\) \(x^2x6=0 (2) x = 3 (3) x = 2\) \(0+32=1\) C 2)The equation x^2 + ax  b = 0 has equal roots \(a^2+4b=0\) one of the roots of the equation x^2 + ax + 15 = 0 is 3 \((a+\sqrt{a^24*15})/2=3\) \(+\sqrt{a^260}=6+a\) \((\sqrt{a^260})^2=(6+a)^2\) \(96=12a\) \(a=8\) \(a^2+4b=0\) \(64+4b=0\) \(b=16\) B 3)We can use some numbers \(2^2+1^2=5=m 2^21^2=5=n\) \(ab=2\) \(\sqrt{m^2n^2}/2 = \sqrt{259}/2 = 2 = ab\) C 4) 3x^2 + 12x 2y^2  12y  39 MAX \(3x^2 + 12x\) has max in (2,12) \(2y^2  12y  39\) has max in (6,3) 3x^2 + 12x 2y^2  12y  39 MAX = 12  3 =9 D
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22 Mar 2013, 06:13
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/nA. 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.
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Re: New Algebra Set!!!
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21 Jul 2013, 22:09
targetgmatchotu 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 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 !! I completed the squares for 3x^2 + 12x  ... and for 2y^2  12y... So, I asked myself what do I need there in order to have (a+b)^2. Hope it's clear.
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Re: New Algebra Set!!!
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Updated on: 18 Mar 2013, 13:34
Edited 1. x^4 = x^3 + 6x^2 => x^2 (x^2  x  6) = 0 The roots of x^2  x  6 are 2 and 3, but 2 cannot be the value of x. So 3 and 0 are the only possible roots. => Sum of all possible solutions = 3. Option D
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Originally posted by GyanOne on 18 Mar 2013, 08:06.
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Re: New Algebra Set!!!
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Updated on: 18 Mar 2013, 10:13
4. 3x^2 + 12x 2y^2  12y  39 = 3(x^2 + 4x  13) + 2(y^26y) Now x^2 +4x  13 has its maximum value at x = 4/2 = 2 and y^2  6y has its maximum value at y=6/2 = 3 Therefore max value of the expression = 3(4 + 8 13) + 2 (9+18) = 27 + 18 = 9 Should be B
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Originally posted by GyanOne on 18 Mar 2013, 08:23.
Last edited by GyanOne on 18 Mar 2013, 10:13, edited 1 time in total.



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Re: New Algebra Set!!!
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18 Mar 2013, 08:40
6. M^2 * n^2 + mn = 12 mn( mn + 1 ) = 12 mn = 3 or mn = 4 ( 3 * 4 = 12; 4 * 3 = 12 ) So m = 3/n or m = 4/n
Answer is E



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Re: New Algebra Set!!!
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18 Mar 2013, 11:55
@Bunuel, for Q5, shouldn't the original question also say that we are only looking for solutions where x is an integer? For this to be satisfied, in x^2 + 2x 15 = m, 4  4(m15)>0 and 44(m15) must be a perfect square => 4 (16m) must be a perfect square and >0 => 16m must be a perfect square and >0 The only values that satisfy this for 10<=m<=10 are m=9,0,7 of which only 7 is positive => probability = 1/3
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Re: New Algebra Set!!!
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