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655-705 Level|   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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Bunuel
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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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Bunuel
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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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