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The next set of medium/hard PS algebra questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:

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.

Hi Bunuel, how come I cannot/should not solve this problem by treating them as if they are two equations and multiple unknowns? As in, x^2 + ax + 15 = 0 ; x^2 +ax -b = 0. Subtract the two equations and you get 15+b=0 so b= -15. Thanks!

Because those are not simultaneous equations, those two are not given to be a system of equations.
_________________

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.

Hello Bunnel,

The solution is excellent. But how do we think in 2 min time, we have to split 39 as -12-18-9. It feels challenging to figure that out. Any pointers how to think ?
_________________

Thank You Very Much, CoolKl Success is the Journey from Knowing to Doing

A Kudo is a gesture, to express the effort helped. Thanks for your Kudos.

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.

Hello Bunnel,

I did the same way as above except when -x^2-2x+15 > 0, so took -ve sign common as x^2+2x-15 < 0 (x+5)(x-3) < 0 x+5 < 0 or x -3 < 0 x < -5 or x < 3. So for me possible values are -6,-7-8,-9,-10. Means 5/21. Any idea where am i wrong ?
_________________

Thank You Very Much, CoolKl Success is the Journey from Knowing to Doing

A Kudo is a gesture, to express the effort helped. Thanks for your Kudos.

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.

Hello Bunnel,

I did the same way as above except when -x^2-2x+15 > 0, so took -ve sign common as x^2+2x-15 < 0 (x+5)(x-3) < 0 x+5 < 0 or x -3 < 0 x < -5 or x < 3. So for me possible values are -6,-7-8,-9,-10. Means 5/21. Any idea where am i wrong ?

When you say that -6,-7-8,-9,-10 satisfy the inequality, did you test any of these values? Ho did you solve (x+5)(x-3) < 0 to get x < -5 or x < 3? What does x < -5 OR x < 3 even mean? It does not make sense... (x+5)(x-3) < 0 holds true for -5 < x < 3.

Hi Bunuel, thank you so much for posting these. Could you please share some Rate and Percent problem solving questions?

Also, if I have exhausted the question bank from GMAT prep (most questions I see are repeating now) and if I have finished Manhattan's 6 tests and the free tests posted on the forum, what resources should I use? I have used the problems listed in your signature, too. My exam is in a week from today. Thank you!

Hi Bunuel, thank you so much for posting these. Could you please share some Rate and Percent problem solving questions?

Also, if I have exhausted the question bank from GMAT prep (most questions I see are repeating now) and if I have finished Manhattan's 6 tests and the free tests posted on the forum, what resources should I use? I have used the problems listed in your signature, too. My exam is in a week from today. Thank you!

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.

Hi, I just solved this equation like this-

x^2+2x-15 = -m

so for m>0; -m<0

x^2+2x-15<0 (x+5)(x-3)<0

x<-5 x<3 .. So I get only 5 solutions.. (-6, -7,-8,-9, -10)

Can you please let me know whats wrong in this.. Thanks in Advance
_________________

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.

Hi, I just solved this equation like this-

x^2+2x-15 = -m

so for m>0; -m<0

x^2+2x-15<0 (x+5)(x-3)<0

x<-5 x<3 .. So I get only 5 solutions.. (-6, -7,-8,-9, -10)

Can you please let me know whats wrong in this.. Thanks in Advance

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.

Hi Bunuel, how come I cannot/should not solve this problem by treating them as if they are two equations and multiple unknowns? As in, x^2 + ax + 15 = 0 ; x^2 +ax -b = 0. Subtract the two equations and you get 15+b=0 so b= -15. Thanks!

Because those are not simultaneous equations, those two are not given to be a system of equations.

Hi Bunuel, sorry if this is a silly question, but how do we know that these are not simultaneous equations? What are the indicators here so I don't treat this as simultaneous equations next time I see a problem such as this. I guess I've been blindly treating problems such as these as a 'multiple equation multiple variable' form of equations and solving them the way you'd solve 2 or 3 equations/2 or 3 unknowns type of problems. Thanks!

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.

You have been clearing doubts , one more need your help. WHy X cannot take negative value? What I understand that If X=-3 then (-3)^4=(3)^4 .

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.

You have been clearing doubts , one more need your help. WHy X cannot take negative value? What I understand that If X=-3 then (-3)^4=(3)^4 .

Please read the whole thread.

\(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)).

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 a positive value on the GMAT.

In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5.
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