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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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18 Mar 2013, 08:04



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18 Mar 2013, 08:06
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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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Last edited by GyanOne on 18 Mar 2013, 13:34, edited 3 times in total.



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18 Mar 2013, 08:11



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Re: New Algebra Set!!! [#permalink]
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18 Mar 2013, 08:17
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2. Let one of the roots be x, then other root is also x. Sum of roots = 2x = a Product of roots = x^2 =  b
For other equation one root is 3 and product of the roots is 15, so other root is 5. Sum of roots = 8 = a or a = 8. From above 2x = 8 or x = 4, b =  x^2 = 16. Answer is B.



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18 Mar 2013, 08:22
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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
Answer is C.



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Re: New Algebra Set!!! [#permalink]
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18 Mar 2013, 08:23
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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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Last edited by GyanOne on 18 Mar 2013, 10:13, edited 1 time in total.



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Re: New Algebra Set!!! [#permalink]
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18 Mar 2013, 08:34
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9. x = (sqrt(5)  sqrt(7))^2 = 5 + 7  2sqrt(35) =~ 12  2*6 = 0 Option A
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Re: New Algebra Set!!! [#permalink]
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18 Mar 2013, 08:40
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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!!! [#permalink]
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18 Mar 2013, 09:02
8. If m is a negative integer and m^3 + 380 = 381m , then what is the value of m? m^3  381m + 380 = 0 Sum of roots = 0 Product of roots = 380 Therefore 19,1,20 is definitely a possible solution set. Sum of roots = 0, Product of roots = 380 B is therefore the right answer.
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Re: New Algebra Set!!! [#permalink]
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18 Mar 2013, 09:03
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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!!! [#permalink]
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18 Mar 2013, 09:04
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 \(x=\sqrt[4]{x^3+6x^2}\) =>\(x^4=x^3+6x^2\) =>\(x^4x^36x^2=0\) Sum of roots for \(ax^n+bx^(n1)+cx^(n2)+... + constant\) = \(b/a\) In the given prob : b= 1, a=1 Sum of roots = \(b/a = (1)/1 = 1\) C
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18 Mar 2013, 09:08
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10. 2x^2  1 = (x+12)^2 2x^2  1 = x^2 + 24x +144 x^2  24x  145 = 0 Product of roots = 145 So answer is A.



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18 Mar 2013, 09:12
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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 \(f(n^2) = 2n^2 1 g(n+12) = (n+12)^2 =>2n^2 1 = (n+12)^2 =>2n^2 1 = n^2 + 24n + 144 => n^2  24n  145 = 0\) Product of all n = product of roots of the above equation. For ax^2 + bx + c = 0 => Product of roors = c/a Product of roots = 145/1 = 145 Ans = A
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18 Mar 2013, 09:26
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7. If\(x^4 = 29x^2  100\), then which of the following is NOT a product of two possible values of x? I. 50 II. 25 III. 100 A. I only B. II only C. III only D. I and II only E. I and III only \(x^4 = 29x^2  100\) \(=> x^429x^2+100 = 0\) \(=> x^4  25x^2  4x^2 + 100=0\) \(=> x^2(x^225)  4(x^225) = 0\) \(=>(x^2  4)(x^225) = 0\) => x = +2,2,+5,5 None of 50, 25 or 100 is possible with product of two possible values of x. what I am missing ?
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Re: New Algebra Set!!! [#permalink]
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Re: New Algebra Set!!! [#permalink]
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18 Mar 2013, 09:33
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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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Re: New Algebra Set!!! [#permalink]
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18 Mar 2013, 09:36
Zarrolou wrote: 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 Better to post each solution as a separate post, since I cannot give more than 1 kudos point for one post.
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Re: New Algebra Set!!! [#permalink]
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18 Mar 2013, 09:37
5. I have taken a longer approach, there should be a better solution. x^2 + 2x 15 = m x^2 + 5x  3x  15 = m (x+5)(x3) = m x will have integer values, since m is also an integer. Putting positive values for x such as ( 1, 2....) we will get values for m which lies between 10 and 10, they are 9, 0 and 7. We will get same result for negative values of x, the equation is a parabola, it will be symmetric Hence the probability is 1/3 Answer is B.



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Re: New Algebra Set!!! [#permalink]
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18 Mar 2013, 09:42
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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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