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



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05 Jun 2013, 07:08
thank you Bunuel,,, I read the question differently ( m raised to 2n raised to 2)



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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"?



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



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Bunuel wrote: 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? 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



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Re: New Algebra Set!!! [#permalink]
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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.
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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=\(3\)



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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.



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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 \(a*b\)\(=\sqrt{(m^2n^2)/4}\)



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Re: New Algebra Set!!! [#permalink]
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20 May 2014, 02:02
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. Why can't we simply deduct the first equation from the second so that, x^2 ans ax gets eliminated and b remains with a value of 15?



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New Algebra Set!!! [#permalink]
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20 May 2014, 02:25
farzana87 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: \(3^3+3a+15=0\). 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. Why can't we simply deduct the first equation from the second so that, x^2 ans ax gets eliminated and b remains with a value of 15? Well, because these equations are not the same: x^28x+16=0 and x^28x+15=0.
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21 Jun 2014, 01:35
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, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power.
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