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New Algebra Set!!! [#permalink]
18 Mar 2013, 06:56
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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:
Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 07:06
1
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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.
Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 07:11
1
This post received KUDOS
Expert's post
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.
Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 07:17
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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.
Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 07:22
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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 = ((m-n)/2)^(1/2)
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
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 _________________
"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks!
Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 08:33
3
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1)\(x^4=x^3+6x^2\) \(x^4-x^3-6x^2=0\) \(x^2(x^2-x-6)=0\) \(x^2=0 (1) x = 0\) \(x^2-x-6=0 (2) x = 3 (3) x = -2\) \(0+3-2=1\) 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 \((-a+-\sqrt{a^2-4*15})/2=3\) \(+-\sqrt{a^2-60}=6+a\) \((\sqrt{a^2-60})^2=(6+a)^2\) \(-96=12a\) \(a=-8\)
\(a^2+4b=0\) \(64+4b=0\) \(b=-16\) B
3)We can use some numbers \(2^2+1^2=5=m 2^2-1^2=5=n\) \(ab=2\)
\(\sqrt{m^2-n^2}/2 = \sqrt{25-9}/2 = 2 = ab\) 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 _________________
It is beyond a doubt that all our knowledge that begins with experience.
Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 08:36
1
This post received KUDOS
Expert's post
Zarrolou wrote:
1)\(x^4=x^3+6x^2\) \(x^4-x^3-6x^2=0\) \(x^2(x^2-x-6)=0\) \(x^2=0 (1) x = 0\) \(x^2-x-6=0 (2) x = 3 (3) x = -2\) \(0+3-2=1\) 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 \((-a+-\sqrt{a^2-4*15})/2=3\) \(+-\sqrt{a^2-60}=6+a\) \((\sqrt{a^2-60})^2=(6+a)^2\) \(-96=12a\) \(a=-8\)
\(a^2+4b=0\) \(64+4b=0\) \(b=-16\) B
3)We can use some numbers \(2^2+1^2=5=m 2^2-1^2=5=n\) \(ab=2\)
\(\sqrt{m^2-n^2}/2 = \sqrt{25-9}/2 = 2 = ab\) 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. _________________
Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 08:37
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)(x-3) = -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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