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New Algebra Set!!!

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Bunuel
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New Algebra Set!!! [#permalink] New post   Updated on: 06 Apr 2023, 00:13
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On the GMAT, algebraic concepts are integral to many questions, and a strong foundation in these areas is essential for success. To perform well, you must be skilled in techniques such as simplification, factoring, and equation solving. The following 10 problems from GMAT Club Tests will challenge your abilities in these areas and help you improve your algebraic aptitude.

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


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


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


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

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


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


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


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


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


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


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


The solutions can be found below on this page.

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Originally posted by Bunuel on 18 Mar 2013, 07:56.
Last edited by Bunuel on 06 Apr 2023, 00:13, edited 4 times in total.
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Re: New Algebra Set!!! [#permalink] New post   22 Mar 2013, 05:47
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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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Re: New Algebra Set!!! [#permalink] New post   22 Mar 2013, 05:01
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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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Re: New Algebra Set!!! [#permalink] New post   22 Mar 2013, 06:30
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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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Re: New Algebra Set!!! [#permalink] New post   22 Mar 2013, 06:08
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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] New post   22 Mar 2013, 05:16
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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] New post   22 Mar 2013, 05:10
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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] New post   22 Mar 2013, 06:44
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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] New post   22 Mar 2013, 06:20
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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] New post   22 Mar 2013, 06:13
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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] New post   22 Mar 2013, 06:35
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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] New post   Updated on: 18 Mar 2013, 13:34
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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, 08:06.
Last edited by GyanOne on 18 Mar 2013, 13:34, edited 3 times in total.
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Re: New Algebra Set!!! [#permalink] New post   18 Mar 2013, 08:11
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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] New post   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 = ((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   Updated on: 18 Mar 2013, 10:13
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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, 08:23.
Last edited by GyanOne on 18 Mar 2013, 10:13, edited 1 time in total.
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Re: New Algebra Set!!! [#permalink] New post   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(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, 09:33
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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, 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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Re: New Algebra Set!!! [#permalink] New post   19 Mar 2013, 02:30
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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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Re: New Algebra Set!!! [#permalink] New post   19 Mar 2013, 10:57
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GyanOne wrote:
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



Hi GyanOne,

Can you please explain how you found x/y which gives max value for the 2 cases here? Is there a formula?
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Re: New Algebra Set!!! [#permalink]
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