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New Algebra Set!!! [#permalink] New post 18 Mar 2013, 06:56
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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:

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

Solution: new-algebra-set-149349-60.html#p1200948


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

Solution: new-algebra-set-149349-60.html#p1200950


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}

Solution: new-algebra-set-149349-60.html#p1200956


4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?

A. -39
B. -9
C. 0
D. 9
E. 39

Solution: new-algebra-set-149349-60.html#p1200962


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

Solution: new-algebra-set-149349-60.html#p1200970


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

Solution: new-algebra-set-149349-60.html#p1200973


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

Solution: new-algebra-set-149349-60.html#p1200975


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

Solution: new-algebra-set-149349-60.html#p1200980


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

Solution: new-algebra-set-149349-60.html#p1200982


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

Solution: new-algebra-set-149349-80.html#p1200987


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Re: New Algebra Set!!! [#permalink] New post 22 May 2013, 21:19
Bunuel wrote:
[
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 thatx>=-6
Hence, x=-2 is a valid root, and sum of all roots should bex=3+(-2) = 1
Please let me know where I am going wrong.

Thanks
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Re: New Algebra Set!!! [#permalink] New post 23 May 2013, 01:00
Expert's post
imhimanshu wrote:
Bunuel wrote:
[
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 thatx>=-6
Hence, x=-2 is a valid root, and sum of all roots should bex=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?
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Re: New Algebra Set!!! [#permalink] New post 23 May 2013, 04:03
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

Solution:

x^2 + ax + 15 = 0 equation has one root as 3. only possible values of getting product of the roots 15 , having one root 3 is 5*3.
x^2-5x-3x+15=0
take common values out => x(x-5)-3(x-5) =0 => x=5 and x=3(what was given) , from this we get a= -8.

x^2 -8x - b =0 => if roots are equal then b^2-4ac=0, 64+4b=0 from this b=-16

Answer
B
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Re: New Algebra Set!!! [#permalink] New post 23 May 2013, 04:42
Expert's post
SathyaNIT wrote:
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

Solution:

By using below substitution values we could reduce the powers of x.

Here values (1,0,-29,0,100) - taken from equation - x^4 = 29x^2 - 100=> (1)x^4+(0)x^3-29x^2+(0)x+100=0
Image



from above explanation we get equation x^2-25=0 => give x=+5,-5

from above calculation we get values of x =+2,-2.

multiplying any of the three roots we have possibility of getting product values, -50,+50.

Ans:

E


The correct answer is B, not E: new-algebra-set-149349-60.html#p1200975

Notice that we are asked "which of the following is NOT a product of three possible values of x".

Hope it helps.
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Re: New Algebra Set!!! [#permalink] New post 25 May 2013, 22:27
Bunuel wrote:
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}

Sum the two equations: 2a^2=m+n;
Subtract the two equations: 2b^2=m-n;

Multiply: 4a^2b^2=m^2-n^2;

Solve for ab: ab=\frac{\sqrt{m^2-n^2}}{2}

Answer: C.


Another method is picking numbers:
let a=2
b=3
m=a^2+b^2 ie 2*2+3*3= 13

Similarly,

n= a^2-b^2 ie 2*2-3*3= -5

Now we have to find a*b ie 2*3=6

plug the values of m=13 and n=-5

Option C gives 6 which we are looking for....

Thus Answer is C..
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Re: New Algebra Set!!! [#permalink] New post 04 Jun 2013, 22:45
In question no.6)
how is m^2n^2=(mn)^2 ? Someone help me.
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Re: New Algebra Set!!! [#permalink] New post 04 Jun 2013, 23:40
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Re: New Algebra Set!!! [#permalink] New post 05 Jun 2013, 06:08
thank you Bunuel,,, I read the question differently ( m raised to 2n raised to 2)
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Re: New Algebra Set!!! [#permalink] New post 22 Jun 2013, 23:35
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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Re: New Algebra Set!!! [#permalink] New post 23 Jun 2013, 00:37
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aquax wrote:
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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Re: New Algebra Set!!! [#permalink] New post 08 Jul 2013, 23:40
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Re: New Algebra Set!!! [#permalink] New post 21 Jul 2013, 21:05
Bunuel wrote:
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 !!

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Re: New Algebra Set!!! [#permalink] New post 10 Oct 2013, 16:57
Bunuel wrote:
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: 3^3+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.



Can you explain what a root of an equation is, as well as a discriminant, I didn't get anything done for this one
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Re: New Algebra Set!!! [#permalink] New post 11 Oct 2013, 01:49
Expert's post
AccipiterQ wrote:
Bunuel wrote:
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: 3^3+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.



Can you explain what a root of an equation is, as well as a discriminant, I didn't get anything done for this one


Check here: http://www.purplemath.com/modules/solvquad4.htm

Hope it helps.
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Re: New Algebra Set!!! [#permalink] New post 24 Oct 2013, 00:01
Expert's post
NeetiGupta wrote:
Bunuel wrote:
Bunuel wrote:
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 thatx>=-6
Hence, x=-2 is a valid root, and sum of all roots should bex=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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Re: New Algebra Set!!! [#permalink] New post 05 Apr 2014, 02:42
Q1.
Option D.
Raising both LHS and RHS to fourth power,and making RHS=0 we get
x^4-x^3-6x^2=0
Taking x^2 common and factorizing,
x^2(x+2)(x-3)=0
Three possible values of x are -2,3 and 0
But x can't be -ve.
So sum of possible values=3
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Re: New Algebra Set!!! [#permalink] New post 05 Apr 2014, 02:47
Q2
Option B.
From second equation,calculate value of a by putting x=3
a=-8
For roots to be equal,D=0
a^2+4b=0
b=-16

Last edited by AKG1593 on 05 Apr 2014, 02:54, edited 1 time in total.
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Re: New Algebra Set!!! [#permalink] New post 05 Apr 2014, 02:54
Q3.
Option C.
a^2+b^2=m
a^2-b^2=n
=>a^2=m+n/2
Or a=\sqrt{(m+n)/2}

Similarly b=\sqrt{(m-n)/2}
Multiplying a*b=\sqrt{(m^2-n^2)/4}
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Re: New Algebra Set!!! [#permalink] New post 20 May 2014, 01:02
Bunuel wrote:
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: 3^3+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.



Why can't we simply deduct the first equation from the second so that, x^2 ans ax gets eliminated and b remains with a value of -15?
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Re: New Algebra Set!!! [#permalink] New post 20 May 2014, 01:25
Expert's post
farzana87 wrote:
Bunuel wrote:
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: 3^3+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.



Why can't we simply deduct the first equation from the second so that, x^2 ans ax gets eliminated and b remains with a value of -15?


Well, because these equation are not the same: x^2-8x+16=0 and x^2-8x+15=0.
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Re: New Algebra Set!!!   [#permalink] 20 May 2014, 01:25
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