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655-705 (Hard)|   Algebra|   Sequences|      
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BelisariusTirto
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One more thing, x cannot be -2, so we don't include it in the sum, but also because in the gmat we do not deal with the square root of negative numbers... Does the fact it's even (4th) root as written in the post matter then?
Bunuel
BelisariusTirto
Hello Bunuel, I was wondering where the x^2 outside the bracket goes in x^2 (x^2-x-6)=0?

In my working it's

x^4-x^3-6x^2=0
then: x^2(x^2-x-6),
x^2(x^2-3x+2x-6)

and then I'm stuck...


Thank you for this post.
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.

It does not go anywhere. We get:

\(x^2(x^2-x-6)=0\);

\(x^2(x^2-3x + 2x -6)=0\);

\(x^2(x(x-3) + 2(x -3))=0\);

\(x^2(x-3)(x + 2)=0\);

\(x^2(x-3)(x+2)=0\).

From this, x^2 gives one of the possible roots for the equation: x = 0.
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Bunuel
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BelisariusTirto
One more thing, x cannot be -2, so we don't include it in the sum, but also because in the gmat we do not deal with the square root of negative numbers... Does the fact it's even (4th) root as written in the post matter then?
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Yes, even roots of negative numbers are not defined on the GMAT; however, that is not the reason we are rejecting x = -2 as a solution. For x = -2, the expression under the 4th root is actually positive (16), not negative, so that’s not the issue. The real issue is that even roots cannot yield negative results! The fourth root of 16 is 2 only—not +2 and -2. Therefore, x = -2 does not satisfy the given equation.
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Hi,

Can you explain the discriminant part further? Where did the "4" in 4b come from?
Bunuel
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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AnimangaStars
Hi,

Can you explain the discriminant part further? Where did the "4" in 4b come from?
Bunuel
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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Check here: https://www.purplemath.com/modules/solvquad4.htm Hope it helps.
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In a parabola, the maximum/minimum value of a function is at the middle point between two zeros
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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?
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But x cannot be negative as it equals to the even (4th) root of some expression - can you pls explain this?

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.
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Bunuel
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saubhikbhaumik
But x cannot be negative as it equals to the even (4th) root of some expression - can you pls explain this?

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.
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This is already addressed in the thread. Please invest some time and read the whole topic. Hope it helps.
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Where did the term "M" disappear? Is it a typo? Ideally, the equation should be simplified by taking "M" to the left had side.

Bunuel
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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findingmyself
Where did the term "M" disappear? Is it a typo? Ideally, the equation should be simplified by taking "M" to the left had side.

Bunuel
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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There is no missing "m." The solution correctly moves all terms to express m = -x^2 - 2x + 15.

The question asks when m is greater than zero, so we solve the inequality -x^2 - 2x + 15 > 0, exactly as shown.
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question 8:
i divide both sides by m since i know it is not equal to zero.
m^2 + 380/m = 381

380/m has to be an integer. now anyone can do it.
Bunuel
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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Hi Bunuel!

Why did you not take -2 and +2 ? Why the or condition considered on for x^2 = 2 and not for x^2 = 25 ? For the latter both -5 and +5 were taken.
Bunuel
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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inciduntveniam
Hi Bunuel!

Why did you not take -2 and +2 ? Why the or condition considered on for x^2 = 2 and not for x^2 = 25 ? For the latter both -5 and +5 were taken.

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Both signs were taken in both cases.

From x^2 = 25, we get x = 5 or x = -5.

From x^2 = 4, we get x = 2 or x = -2.

So the four possible values are 5, -5, 2, and -2. The solution already uses all four.
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Hi inciduntveniam,

Great catch, and the fix is to notice that Q1 and Q7 are two different situations, even though both involve even powers.

Q7 (where ±5 and ±2 are both kept): the equation is x4 = 29x2 - 100. There is no root sign here. When you solve it you reach x2 = 25 and x2 = 4. Solving x2 = 25 means asking "what values of x, when squared, give 25?" - and that is both +5 and -5. Same for x2 = 4, giving +2 and -2. So all four values are legitimate. Nothing forces x to be positive.

Q1 (where -2 is thrown out): the equation is x = 4√(x3 + 6x2). Look at the left side: x is set equal to a 4th-root expression. A root sign like √ or 4√ is an operation that only ever outputs a non-negative number (4√16 = 2, never -2). So whatever x equals, it must be ≥ 0. That's why the candidate x = -2 fails - plug it in and you get 4√16 = 2, so you'd be claiming -2 = 2, which is false.

The one distinction to hold onto:
- x2 = 25 - two answers, +5 and -5 (you're solving an equation).
- x = √25 - one answer, +5 (you're reading the output of a root sign, which is never negative).

Q7 is the first case. Q1 is the second case - x is literally equal to a root, so it inherits the root's non-negative restriction.

Quick self-test to lock it in:
- How many solutions does y2 = 9 have? (Two: +3 and -3.)
- What is √9? (Just 3.)

Same numbers, different question - and that difference is exactly what separates Q1 from Q7.

Answer: D

inciduntveniam
Hi Bunuel!

Why did you not take -2 and +2 ? Why the or condition considered on for x^2 = 2 and not for x^2 = 25 ? For the latter both -5 and +5 were taken.

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