<a class="topic-title" href="https://gmatclub.com/forum/new-algebra-set-149349.html"> New Algebra Set!!!

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Originally posted by Bunuel on 18 Mar 2013, 06:56.
Last edited by Bunuel on 05 Apr 2023, 23:13, edited 4 times in total.

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Bunuel

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

22 Mar 2013, 04:47

Kudos

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4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?

A. -39
B. -9
C. 0
D. 9
E. 39

\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=-3(x^2-4x+4)-2(y^2+6y+9)-9=\)
\(=-3(x-2)^2-2(y+3)^2-9\).

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, then the maximum value of the whole expression is \(0+0-9=-9\).

Answer: B.
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Bunuel

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

22 Mar 2013, 04:01

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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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Bunuel

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

22 Mar 2013, 05:30

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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

Given \(m^3 + 380 = 380m+m\).

Re-arrange: \(m^3-m= 380m-380\).

\(m(m+1)(m-1)=380(m-1)\). Since m is a negative integer, then \(m-1\neq{0}\) and we can safely reduce by \(m-1\) to get \(m(m+1)=380\).

So, we have that 380 is the product of two consecutive negative integers: \(380=-20*(-19)\), hence \(m=-20\).

Answer: B.
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Bunuel

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

22 Mar 2013, 05:08

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

22 Mar 2013, 04:16

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3. If \(a\) and \(b\) are positive numbers, such that \(a^2 + b^2 = m\) and \(a^2 - b^2 = n\), what is \(ab\) in terms of \(m\) and \(n\)?

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. \(\frac{\sqrt{m^2+n^2} }{2}\)

By summing the two equations, we get: \(2a^2 = m + n\).

By subtracting the second equation from the first, we obtain: \(2b^2 = m - n\).

Multiplying these two results, we have: \(4a^2b^2 = m^2 - n^2\).

From this, solving for \(ab\) yields: \(ab = \frac{\sqrt{m^2 - n^2} }{2}\).

Answer: C.
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Re: New Algebra Set!!! [#permalink]

22 Mar 2013, 04:10

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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

Since one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3, then substituting we'll get: \(3^2+3a+15=0\). Solving for \(a\) gives \(a=-8\).

Substitute \(a=-8\) in the first equation: \(x^2-8x-b=0\).

Now, we know that it has equal roots thus its discriminant must equal to zero: \(d=(-8)^2+4b=0\). Solving for \(b\) gives \(b=-16\).

Answer: B.
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Re: New Algebra Set!!! [#permalink]

22 Mar 2013, 05:44

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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

\(f(x) = 2x - 1\), hence \(f(n^2)=2n^2-1\).
\(g(x) = x^2\), hence \(g(n+12)=(n+12)^2=n^2+24n+144\).

Since given that \(f(n^2)=g(n+12)\), then \(2n^2-1=n^2+24n+144\). Re-arranging gives \(n^2-24n-145=0\).

Next, Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(n_1*n_2=-145\).

Answer: A.
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Re: New Algebra Set!!! [#permalink]

22 Mar 2013, 05:20

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

22 Mar 2013, 05:13

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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

Rearrange: \((mn)^2 + mn - 12=0\).

Factorize for \(mn\): \((mn+4)(mn-3)=0\). Thus, \(mn=-4\) or \(mn=3\).

Therefore, we find \(m=-\frac{4}{n}\) or \(m=\frac{3}{n}\).

Answer: E
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Re: New Algebra Set!!! [#permalink]

22 Mar 2013, 05:35

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

Updated on: 18 Mar 2013, 12:34

Kudos

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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

Originally posted by GyanOne on 18 Mar 2013, 07:06.
Last edited by GyanOne on 18 Mar 2013, 12:34, edited 3 times in total.

Bunuel

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

18 Mar 2013, 07:11

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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]

18 Mar 2013, 07:22

Kudos

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]

Updated on: 18 Mar 2013, 09:13

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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

Originally posted by GyanOne on 18 Mar 2013, 07:23.
Last edited by GyanOne on 18 Mar 2013, 09:13, edited 1 time in total.

Abhii46

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

18 Mar 2013, 08:03

Kudos

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]

18 Mar 2013, 08:33

Kudos

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]

18 Mar 2013, 08:42

Kudos

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]

19 Mar 2013, 01:30

Kudos

7. \(x^4 = 29x^2 - 100\)

or \(x^4 - 29x^2+100 = 0\)

Let \(x^2 = t\), thus

\(t^2-29+100 = 0\)

or (t-25)(t-4) = 0

t = 25 --> x = 5/-5

or

t = 4 --> x = 2/-2

Out of any three possible values of x, only 25 is not possible.

B.
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