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New Algebra Set!!!

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New Algebra Set!!! [#permalink] New post 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:

A. -2
B. 0
C. 1
D. 3
E. 5

Solution: new-algebra-set-149349-60.html#p1200948


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: new-algebra-set-149349-60.html#p1200950


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{m-n}}{2}
B. \frac{\sqrt{mn}}{2}
C. \frac{\sqrt{m^2-n^2}}{2}
D. \frac{\sqrt{n^2-m^2}}{2}
E. \frac{\sqrt{m^2+n^2}}{2}

Solution: new-algebra-set-149349-60.html#p1200956


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: new-algebra-set-149349-60.html#p1200962


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: new-algebra-set-149349-60.html#p1200970


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: new-algebra-set-149349-60.html#p1200973


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: new-algebra-set-149349-60.html#p1200975


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: new-algebra-set-149349-60.html#p1200980


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: new-algebra-set-149349-60.html#p1200982


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: new-algebra-set-149349-80.html#p1200987


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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 07:04
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 07:06
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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, 12:34, edited 3 times in total.
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 07:11
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.

Option C


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

ab = ((m^2 - n^2)^(1/2))/2

Answer is C.
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 07:23
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4. -3x^2 + 12x -2y^2 - 12y - 39 = 3(-x^2 + 4x - 13) + 2(-y^2-6y)

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, 09:13, edited 1 time in total.
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 07:34
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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] New post 18 Mar 2013, 07:40
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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] New post 18 Mar 2013, 08:02
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] New post 18 Mar 2013, 08:03
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8. m^3 + 380 = 381m
m^3 + 380 = 380m + m
m^3 -m = 380m - 380
m(m-1)(m+1) = 380 (m-1)
m(m+1) = 380 ( -20 * -19; m is negative )
m = -20.

Answer is B
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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:04
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^4-x^3-6x^2=0

Sum of roots for ax^n+bx^(n-1)+cx^(n-2)+... + constant = -b/a
In the given prob : b= -1, a=1

Sum of roots = -b/a = -(-1)/1 = 1
C

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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:08
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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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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:12
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10.Iff(x) = 2x - 1 andg(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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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:26
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7. Ifx^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^4-29x^2+100 = 0
=> x^4 - 25x^2 - 4x^2 + 100=0
=> x^2(x^2-25) - 4(x^2-25) = 0
=>(x^2 - 4)(x^2-25) = 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] New post 18 Mar 2013, 08:32
Expert's post
ConnectTheDots wrote:
7. Ifx^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^4-29x^2+100 = 0
=> x^4 - 25x^2 - 4x^2 + 100=0
=> x^2(x^2-25) - 4(x^2-25) = 0
=>(x^2 - 4)(x^2-25) = 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 ?


Edited typos. Thank you.

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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:33
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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

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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:36
1
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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


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Re: New Algebra Set!!! [#permalink] New post 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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Re: New Algebra Set!!! [#permalink] New post 18 Mar 2013, 08:42
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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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Re: New Algebra Set!!!   [#permalink] 18 Mar 2013, 08:42
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