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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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25 Mar 2013, 12:31
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, Query here. First of all we don't necessarily have to solve for all the three roots. Since, it is asking for a composite value i.e. the sum, we can directly obtain it. We can rearrange to get: \(x^4x^36x^2=0\) Over here, by the system of poly equations Sum of roots =  (\(coeff ofx^3/coeff of x^4\)) Similarly, Sum of roots taken two at a time = (\(coeff ofx^2/coeff of x^4\)) Product of roots = "Zero" as there is no constant term. Also, 2 satisfies the equation \(x^4x^36x^2=0\) So, shouldn't the answer be 1, with roots as 0,2,3?



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 02:07
Sinner1706 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, Query here. First of all we don't necessarily have to solve for all the three roots. Since, it is asking for a composite value i.e. the sum, we can directly obtain it. We can rearrange to get: \(x^4x^36x^2=0\) Over here, by the system of poly equations Sum of roots =  (\(coeff ofx^3/coeff of x^4\)) Similarly, Sum of roots taken two at a time = (\(coeff ofx^2/coeff of x^4\)) Product of roots = "Zero" as there is no constant term. Also, 2 satisfies the equation \(x^4x^36x^2=0\) So, shouldn't the answer be 1, with roots as 0,2,3? It's explained in the solution above: \(x\) 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 \(x=3\).
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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:03
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
Kudos points for each correct solution!!![/quote]
X^4 = X^3 + 6x^2.Solving this we get teh values of x as 0, 3 , 2. But X cannot be negative as X is the positive fourth root of an expression(\sqrt{Expression}>= 0) So X = 0 and 3 Sum of the values = 0+3 = 3



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:12
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 Kudos points for each correct solution!!![/quote]
Since 3 is the root of the equation x^2 + ax + 15 = 0, therefore 3 must satisfy the equation x^2 + ax + 15 = 0 . On putting x=3 in the equation x^2 + ax + 15 = 0 , we get a= 8 On putting the value of a in the original equation x^2 + ax  b = 0, we get x^2  8x  b = 0. As per question x^2  8x  b = 0 has equal roots so Discriminant(D) = b^2 4ac = 0 = (8)^2  4.1.(b)= 0. From this we get the value of b as 16



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:17
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. \(\frac{\sqrt{m^2+n^2}}{2}\) Kudos points for each correct solution!!![/quote]
Adding the given two equations we get 2a^2 = (m+n)/2 Subtracting the given two equations we get 2b^2 = (mn)/2
Multiplying above two equations we get 4a^2b^2 = (m^2  n^2) => ab = \(\frac{\sqrt{m^2n^2}}{2}\)



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:29
4. What is the maximum value of 3x^2 + 12x 2y^2  12y  39 ?
A. 39 B. 9 C. 0 D. 9 E. 39 Kudos points for each correct solution!!![/quote]
Rearranging the given equation we get 3(x^2  4x + 4) 2(y^2 + 6y +9)  9 = 3(x 2)^2  2(y +3)^2 9 So, we need to maximize the value of 3(x 2)^2  2(y +3)^2 9
Since, the maximum value of 3(x 2)^2 and  2(y +3)^2 is zero when x= 2 and y = 3, then the maximum value of the whole expression is 0 + 0 9 = 9



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:39
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 Kudos points for each correct solution!!![/quote]
On Rearranging the given equation we get m =  x^2  2x +15
For m to be positive  x^2  2x +15 should be greater then 0. Solving we get (x + 5)(x  3)<0. Since x is an integer we get the values of x as 4 , 3, 2, 1, 0, 1, 2 (total 7 values)
Given that X is an integer from 10 and 10, inclusive (21 values)
So, the probability is 7/21=1/3.
Answer: B.



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:44
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 Kudos points for each correct solution!!![/quote]
Rearranging we get m^2n^2 + mn  12 = 0 => (mn + 4)(mn  3)= 0 =>mn = 4 or mn = 3 => m= 4/n or 3/n E. I and III only



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 05:55
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 Kudos points for each correct solution!!![/quote]
Rearranging the equation we get x^4  29x^2 + 100 = 0 => (x^2  25)(x^  4) =0 => x= +5, 5, +2, 2 From these values of x we can get 50 (5 * +5 * +2), +50(5 * +5 * 2) but we can't get 25
B. II only



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 06:17
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
on rearranging we get m^3 381m = 380 => m(m^2  381)= 380 On checking the options and hit and trial, we get the values of m to be 1 and 20. But since m is negative so m = 20
B. 20



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 06:29
9. If \(x=(\sqrt{5}\sqrt{7})^2\), then the best approximation of x is:
A. 0 B. 1 C. 2 D. 3 E. 4
\(x=(\sqrt{5}\sqrt{7})^2\)
x= 5 + 7  2. \(\sqrt{35}\) \(\sqrt{35}\) ~ 6
x = 5+7  2*6 = 12  12 = 0



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Re: New Algebra Set!!! [#permalink]
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26 Mar 2013, 06:34
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(n^2) = \(2.n^2  1\) g(n + 12) = (n + 12)^2 f(n^2) = g(n + 12) => \(2.n^2  1\) = (n + 12)^2 => 2.n^2 1 = n^2 + 144 +24.n => n^2 24n 145 = 0 From the above equation we get n = 5 and 29 So product of values of n = 5 * 29 = 145
A. 145



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Re: New Algebra Set!!! [#permalink]
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28 Mar 2013, 01:23
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#p1200948 x^4=x^3 + 6x^2 x^4  x^3  6x^2 = 0 > x^2 (x^2  x  6) = 0 > x^2 (x3)(x+2) = 0 x=0 or 3 or 2. With x = 2 original equation does not hold true so possible values for X are 3 and 0 Hence their sum is 3 Choice D. 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 Solution: newalgebraset14934960.html#p1200950 x^2 + ax  b = 0 that means a^2  4(b) = 0 > a^2 + 4b = 0 Equation I x^2 + ax + 15 has 3 as a root, so 9 + 3a + 15 = 0 > 3a + 24 = 0 > a = 8 We will put the value of a in equation I > a^2 + 4b = 0 > 64 + 4b = 0 > b = 16 > Choice B is the answer.
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28 Mar 2013, 01:37
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} Solution: newalgebraset14934960.html#p1200956 a^2 + b^2 = m a^2  b^2 = n  2a^2 = m+n > a^2 = (m+n)/2 > a = sq.root (m+n) / sq.root 2 similarly b = sq.root (mn) / sq.root 2 So ab = (sq.root (m+n) / sq.root 2)(sq.root (mn) / sq.root 2) > ab = sq.root(m^2  n^2)/2Using (a+b)(ab) = a^2  b^2 Choice C 4. 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#p1200962
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Re: New Algebra Set!!! [#permalink]
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28 Mar 2013, 01:49
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 Solution: newalgebraset14934960.html#p1200970 there are 21 possible values of x so does m has 21 possible values from x = 4 to x = 2, m gives negative value; Total 7 values so the probability m being negative is 7/21> 1/3 > choice B 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 Solution: newalgebraset14934960.html#p1200973 m^2n^2 + mn = 12 > mn(mn+1) = 12 > here mn and mn+1 are consecutive integers. So 12 is basically a product of 2 consecutive intergers those integers must be 3 and 4 or 3 and 4 so possible values for mn are mn = 3 and mn+1 = 4 > m = (3/n) mn = 4 and mn + 1 =3 > m =(4/n) Choice E
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28 Mar 2013, 04:05
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: newalgebraset14934960.html#p1200975 x^4  29x^2 + 100 = 0 > let X^2 be y > y^2  29y + 100 = 0 > y^2  25y  4y + 100 = 0 > (y25)(y4)=0 y=25 > x^2 = 25 > x= 5 or 5 y=4> x^2 = 4 > x = 2 or 2 Possible values of x are 5, 5, 2, and 2 Out of options 25 can never be a product of any 3 values of x, so Choice B is the answer 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 Solution: newalgebraset14934960.html#p1200980 This was a bit complex one m^3  381m =  380 > m(m^2  381) =  380 Since 380 is the product of m(m^2  381), any one of either m or (m^2  381) must be negative. Since we are given that m is a negative integer, then m^2  381 must be positive So m^2  381 > 0 > m^2 > 381 > m > 19.5 approx. so m > 19.5if m is positive and m < 19.5  if m is negative We know that m is negative so m must equal to 20 : choice B
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28 Mar 2013, 04:10
9. 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#p1200982 x=(sq.root 5  sq.root 7)^2 > x = 5 + 7 2(sq.root 35) using (ab)^2 = a^2 + b^2  2ab 12  2(sq.root 35) > assuming sq.root 35 = 6 > 1212 = 0 Choice A 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 Solution: newalgebraset14934980.html#p1200987
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Re: New Algebra Set!!! [#permalink]
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29 Mar 2013, 06:54
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, Could you please explain why 2 is not a solution? Is it because x is a nonnegative number? Because if I plug in the value of 2 to the equation \(x^4=x^3+6x^2\) I get \(2=\sqrt[4]{16}\) which can be true isn't it, as \(\sqrt[4]{16}=2\) Also could you explain why even root of an expression cannot be negative when we have an expression which is a square, because we know that \(\sqrt{{x^2}}=x\)? Thank you



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Re: New Algebra Set!!! [#permalink]
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11 Apr 2013, 15:31
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. I still cannot get how did you rearrange the given expression, bunuel . Could you please elaborate more ? Thanks in advance
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Re: New Algebra Set!!! [#permalink]
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12 Apr 2013, 04:29
Bunuel wrote: \(m^2n^2 + mn = 12\) > \(m^2n^2 + mn 12=0\) > \((mn)^2 + mn  12=0\) > say mn=x, so we have \(x^2+x12=0\) > \((x+4)(x3)=0\) > \(x=4\) or \(x=3\) > \(mn=4\) or \(mn=3\). Solving and Factoring Quadratics: http://www.purplemath.com/modules/solvquad.htmhttp://www.purplemath.com/modules/factquad.htmHope it helps. ohhhhhhhhhhhhhhhhhhhhh I thought that m is raised to the power 2n and the whole bracket raised to the power 2 ...that is what made me confused . Thanks a million , Bunuel
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Re: New Algebra Set!!!
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12 Apr 2013, 04:29



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