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 {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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New Algebra Set!!!

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souvik101990

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19 Mar 2013, 11:18

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GyanOne

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?

$-3x^2 + 12x -2y^2 - 12y - 39$

$=-3(x^2-4x)-2(y^2+6y) -39$

$=-3(x^2-4x+4-4)-2(y^2+6y+9-9)-39$

$=-3(x-2)^2 +12 -2(y+3)^2 +18 -39$

NOW The above expression will be minimum IFF $(x-2)^2=0$ and $(y+3)^2=0$

which leaves the maximum value to be $12+18-39=-9$

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19 Mar 2013, 17:22

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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

Sol: The given eqn can be written as mn (mn + 1) = 12
Now mn, mn+ will have to be consecutive Integers and possible combinations are

mn=3 and mn+ 1= 4 ------> possible value of m will be 3/n
or mn+1 = -4 and mn= -3 -------. possible value of m will be -3/n

Ans should be III only i.e Option C

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19 Mar 2013, 17:32

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5. If x^2 + 2x -15 = -m, where m 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

This was little trickier but here is my approach

We need to know Probability that m is greater than 0.
Integer is from -10 to 10 inclusive.
Consider value of m greater than 0 only and put in the value in the given eqn

For ex let us say m= 2, the eqn becomes
x^2+2x -15= -2----. x^2+ 2x -13=0 -----> Eqn will have roots of the form -b+/- \sqrt{b^2-4ac}/ 2

We see that for m= 7, we get the eqn as x^2+ 2x -8 =0, Solving for x we get,
x = 4 or -2.

This seems to be the only case where the eqn gives us 2 real roots and therefore looking at option choices selected B (assuming other 2 cases will be for value of m between -10 to 0)

Ans Choice B

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19 Mar 2013, 17:39

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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

Sol: Adding the 2 equations we get a= \sqrt{(m+n)/2}
Subtracting Eqn 2 from 1, we get b= \sqrt{m-n/2}

Ans Option 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

Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get

3^2 +3a+15=0----> Solving for a we get a=-8.

Putting the value if a in Eqn 1 we get
x^2 -8x-b=0
We know for Eqn ax^2 +bx+c=-0
Sum of roots is given by -b/a
Product of roots is c/a

Therefore from Eqn1 (after subsitution of value of a) we get
Sum of roots = -8 and roots are equal ie. -4 and -4
Product of roots will be 16. Therefore Option 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, then ab in terms of m and n equals to:

A.
B.
C.
D.
E.

Soln: Take equation 2, square on both sides, substitute for a^2 + b^2 with m, Ans (C)

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

Soln: Used a 'gestimation' method. Since the unit digit of the resulting values on both sides has to be same, the only number from the options that will fit this condition is -20. Ans (B)

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1. If x = (x^3 + 6x^2)^1/4, then the sum of all possible solutions for x is:

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

My take ==> C

Approach:

x = (x^3 + 6x^2)^1/4
x^4 = x^3 + 6x^2
x^4 - x^3 - 6x^2 = 0
x^2(x^2 - x - 6) = 0
x^2 * (x-3) * (x+2) = 0

So Sum of all the possible solution of x = 1

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7 - B

x^4=29x^2-100
x^4-29x^2+100=0
which is equal to

(x-2)(x+2)(x-5)(x+5) = 0

roots are 2,-2,5,-5

thus B is the right answer ==> 25 can't be the result of the multiplication of any three roots.

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Hi Bunuel,
Thanks for the challenging ones.

Quote:

Given $m^3 + 380 = 380m+m$.

I was able to backsolve this question, as I couldn't able to figure out the above step. Can you please post your thinking pattern behind this question. I mean, how did you think about breaking the term 381m into 380m + m --->(A)
Any particular indicators in the question that make you do so.
I'm sure during exam time, I will not be able to do so.

Quote:

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.

I have read somewhere that cancelling the like terms in GMAT is dangerous. I can sense that in your post as well( Red colored part). Request you to please enlighten me with this concept as why did you need to make extra caution while cancelling (m-1) terms, instead of cancelling it without given a thought.

Thanks
H

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imhimanshu

Quote:

Consider equation xy=x. Here we cannot reduce by x, because x could be zero and we cannot divide by zero. What we can do is: xy=x --> xy-x=0 --> x(y-1)=0 --> x=0 or y=1.

Now, in the question we have $m(m+1)(m-1)=380(m-1)$ and we know that m is a negative number, thus m-1 does not equal to 0, which means that we can safely reduce by m-1.

Hope it's clear.

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Bunuel

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

Re-arrange: $(mn)^2 + mn - 12=0$.

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

So, we have that $m=-\frac{4}{n}$ or $m=\frac{3}{n}$.

Answer: E.

I still cannot get how did you rearrange the given expression, bunuel . Could you please elaborate more ? Thanks in advance

Bunuel

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Bunuel

I still cannot get how did you rearrange the given expression, bunuel . Could you please elaborate more ? Thanks in advance

$m^2n^2 + mn = 12$ --> $m^2n^2 + mn -12=0$ --> $(mn)^2 + mn - 12=0$ --> say mn=x, so we have $x^2+x-12=0$ --> $(x+4)(x-3)=0$ --> $x=-4$ or $x=3$ --> $mn=-4$ or $mn=3$.

Solving and Factoring Quadratics:
https://www.purplemath.com/modules/solvquad.htm
https://www.purplemath.com/modules/factquad.htm

Hope it helps.

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Hi,

For the first question, what do you mean by "But X cannot be negative as it equals to the even (4th) root of some expression"?

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aquax

Hi,

For the first question, what do you mean by "But X cannot be negative as it equals to the even (4th) root of some expression"?

$x$ cannot be negative as it equals to the even (4th) root of some expression ($\sqrt[even]{expression}\geq{0}$), thus only two solution are valid $x=0$ and $x=3$.

When the GMAT provides the square root sign for an even root, such as $\sqrt{x}$ or $\sqrt[4]{x}$, then the only accepted answer is the positive root. That is, $\sqrt{25}=5$, NOT +5 or -5. Even roots have only non-negative value on the GMAT.

In contrast, the equation $x^2=25$ has TWO solutions, $\sqrt{25}=+5$ and $-\sqrt{25}=-5$.

Hope it helps.

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Bunuel

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.

Hi Bunuel,

Just as a matter of fact can you please explain that how you reached to factorization. In other words how a candidate will decide how to split -39 between (X,Y).

Your reply is appreciated !!

Rgds,
TGC !!

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Bunuel

I completed the squares for -3x^2 + 12x - ... and for -2y^2 - 12y-... So, I asked myself what do I need there in order to have (a+b)^2.

Hope it's clear.

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Bunuel

imhimanshu

Bunuel

[
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.

Hi Bunuel,
Request you to please let me know where I'm making a mistake.

Since $x^3+6*x^2 >= 0$
Then, $x^2(x+6)>=0$

i.e $(x-0)^2(x-(-6))>=0$
This implies that$x>=-6$
Hence, x=-2 is a valid root, and sum of all roots should be$x=3+(-2) = 1$
Please let me know where I am going wrong.

Thanks

Plug x=-2 into $x=\sqrt[4]{x^3+6x^2}$. Does the equation hold true?

Hi Brunel if say x=16.
x^1/2 has two values +4 & -4.
So why x^1/4 cannot have +2 & -2 as as values ?
Please explain

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Bunuel

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.

Hi Bunuel,
Request you to please let me know where I'm making a mistake.

Since $x^3+6*x^2 >= 0$
Then, $x^2(x+6)>=0$

i.e $(x-0)^2(x-(-6))>=0$
This implies that$x>=-6$
Hence, x=-2 is a valid root, and sum of all roots should be$x=3+(-2) = 1$
Please let me know where I am going wrong.

Thanks

Plug x=-2 into $x=\sqrt[4]{x^3+6x^2}$. Does the equation hold true?

Hi Brunel if say x=16.
x^1/2 has two values +4 & -4.
So why x^1/4 cannot have +2 & -2 as as values ?
Please explain

First of all it's 4th root not 2nd root. Next, $\sqrt[4]{16}=2$, not +2 or -2.

When the GMAT provides the square root sign for an even root, such as $\sqrt{x}$ or $\sqrt[4]{x}$, then the only accepted answer is the positive root. That is, $\sqrt{25}=5$, NOT +5 or -5.

In contrast, the equation $x^2=25$ has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Hope it helps.

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Bunuel

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.

Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power.