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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.
_________________
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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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.
_________________
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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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.
_________________
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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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.
_________________
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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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.
_________________
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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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.
_________________
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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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.
_________________
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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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.
_________________
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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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.
_________________
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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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
_________________
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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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.
_________________
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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General Discussion
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
_________________
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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Some questions are tricky!!!
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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.
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.
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
_________________
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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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
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
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
\(x^4-x^3-6x^2=0\)
\(x^2(x^2-x-6)=0\)
\(x^2=0 (1) x = 0\)
\(x^2-x-6=0 (2) x = 3 (3) x = -2\)
\(0+3-2=1\)
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
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
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
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.
_________________
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.
_________________
GyanOne
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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