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New Algebra Set!!! [#permalink]
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18 Mar 2013, 07:56
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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!!!
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Re: New Algebra Set!!! [#permalink]
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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



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Re: New Algebra Set!!! [#permalink]
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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.
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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.
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Re: New Algebra Set!!! [#permalink]
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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!



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Re: New Algebra Set!!! [#permalink]
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06 Feb 2018, 10:44



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Re: New Algebra Set!!! [#permalink]
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06 Feb 2018, 10:56
Bunuel wrote: 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. ok, thank you very much!



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Re: New Algebra Set!!! [#permalink]
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17 Feb 2018, 05:19
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 Can u pls elaborate on d=(8)^2+4b? Thanks in advance



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Re: New Algebra Set!!! [#permalink]
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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 ?



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Re: New Algebra Set!!! [#permalink]
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22 Feb 2018, 22:53



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Re: New Algebra Set!!! [#permalink]
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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.



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New Algebra Set!!! [#permalink]
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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: \(g(n+12)=(n+12)^2=n^2+24n+144\).
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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.
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