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


Kudos points for each correct solution!!!
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OA's and solutions will be published on Friday.
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10.
f(n^2) = g(n+12)
2n^2 - 1 = (n+12)^2
2n^2 - 1 = n^2 + 24n + 144
n^2 - 24n - 145 = 0
(n-29)(n+5)=0
n= 29 or -5
29*(-5) = -145 - Answer A

1.c
2.b
3.c
4.b
5. dont understand the question
6. e
7. b
8. e (guess)
9. a
10. e

:- for 8 ,it is B .. Put thr value ,-20 will satisfy

I agree. Proper answer to 5 as written is 10/21

If x is not explicitly constrained or if x is not constrained by the equation given, then we have to respect all real values of x, integer or not. Since x is not constrained to integers only, we have to respect the possibility that x is not an integer.

Since -10 <= m <= 10 and the equation is x^2 + 2x - 15 = -m, we will always have x^2 + 2x + c = 0 where -25 <= c <= -5. Using quadratic formula (just to prove it, though you won't need to know it for the GMAT), we need the calculate the discriminant, which is b^2 - 4ac, to determine whether or not there are real values for x. Here we get 2^2 - 4*1*c. Since c is always negative, we will always have a positive number for the discriminant. This means that we will always have a positive number underneath the square root of the quadratic formula and, therefore, 2 real outcomes for x.

There are 21 integer values of m (10 positive, 10 negative, and 0) and 10 of them are positive, so 10/21.

Good question though if we constrain x to only integer values.

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;

Re-arrange and factor out x^2: x^2(x^2-x-6)=0;

Factorize: x^2(x-3)(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.
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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}

Sum the two equations: 2a^2=m+n;
Subtract the two equations: 2b^2=m-n;

Multiply: 4a^2b^2=m^2-n^2;

Solve for ab: ab=\frac{\sqrt{m^2-n^2}}{2}

Answer: C.
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Sorry, there was a typo in the stem .

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

Re-arrange the given equation: -x^2-2x+15=m.

Given that x is an integer from -10 and 10, inclusive (21 values) we need to find the probability that -x^2-2x+15 is greater than zero, so the probability that -x^2-2x+15>0.

Factorize: (x+5)(3-x)>0. This equation holds true for -5<x<3.

Since x is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

So, the probability is 7/21=1/3.

Answer: B.
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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

Re-arrange and factor for x^2: (x^2-25)(x^2-4)=0.

So, we have that x=5, x=-5, x=2, or x=-2.

-50=5*(-5)*2;
50=5*(-5)*(-2).

Only 25 is NOT a product of three possible values of x

Answer: B.
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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=5-2\sqrt{35}+7=12-2\sqrt{35}.

Since \sqrt{35}\approx{6}, then 12-2\sqrt{35}\approx{12-2*6}=0.

Answer: A.
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