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New Algebra Set!!!

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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]

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New post 12 Apr 2013, 05:25
Narenn wrote:
nave81 wrote:
Bunuel wrote:
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.



Hi Bunuel,

Could you please explain why -2 is not a solution? Is it because x is a non-negative number? Because if I plug in the value of -2 to the equation \(x^4=x^3+6x^2\) I get \(-2=\sqrt[4]{16}\) which can be true isn't it, as \(\sqrt[4]{16}=|2|\)

Also could you explain why even root of an expression cannot be negative when we have an expression which is a square, because we know that \(\sqrt{{x^2}}=|x|\)?

Thank you


Note the Important Difference.

On the GMAT if X^2 = 4 then x = +/- 2 or |x| = 2 However if x= sq.root(4) then x has to be positive i.e. 2 and it can not take negative value.
when we plug -2 as the value of x in the equation we would get -2 = 4th root of 16 -----> -2 = 2 This is because fourth root of 16 is 2 and not -2

The rule is even root of a number can not be negative on the GMAT

Regards,

Abhijit.


Hey Abhijit,

This is a very interesting point that you have made here. This statement that -2 is the fourth root of 16 is admissible under normal circumstances but is not in a GMAT question. Can you mention a source which would list out such peculiar rules for GMAT quants?

Thanks in advance

Anshuman
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Re: New Algebra Set!!! [#permalink]

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New post 13 Apr 2013, 09:01
Hi Anshuman,

For this you need to go thru some strategy books of GMAT Quant.
In My Opinion, two important sources are,
1) Manhattan GMAT Quant strategy guides 4th or 5th Ed
2) GMAT Club Math Book - This is free and you can download it from this website free of cost (Check BB's profile)
Furthermore you can check my signature. There i have attached my articles prepared on some peculiar topics. You may find them useful.

Thanks,

Abhijit
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Re: New Algebra Set!!! [#permalink]

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New post 22 May 2013, 22:19
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 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
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Re: New Algebra Set!!! [#permalink]

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New post 23 May 2013, 02:00
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 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?
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Re: New Algebra Set!!! [#permalink]

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New post 23 May 2013, 05:03
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]

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New post 23 May 2013, 05:42
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]

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New post 25 May 2013, 23:27
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]

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New post 04 Jun 2013, 23:45
In question no.6)
how is m^2n^2=(mn)^2 ? Someone help me.
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Re: New Algebra Set!!! [#permalink]

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New post 05 Jun 2013, 00:40
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Re: New Algebra Set!!! [#permalink]

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New post 05 Jun 2013, 07:08
thank you Bunuel,,, I read the question differently ( m raised to 2n raised to 2)
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Re: New Algebra Set!!! [#permalink]

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New post 23 Jun 2013, 00:35
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]

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New post 23 Jun 2013, 01:37
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]

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Re: New Algebra Set!!! [#permalink]

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New post 21 Jul 2013, 22:05
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]

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New post 10 Oct 2013, 17:57
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]

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New post 11 Oct 2013, 02:49
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]

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New post 24 Oct 2013, 01:01
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 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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Re: New Algebra Set!!! [#permalink]

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New post 05 Apr 2014, 03:42
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]

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New post 05 Apr 2014, 03:47
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, 03:54, edited 1 time in total.
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Re: New Algebra Set!!! [#permalink]

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New post 05 Apr 2014, 03:54
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 \(a*b\)\(=\sqrt{(m^2-n^2)/4}\)
Re: New Algebra Set!!!   [#permalink] 05 Apr 2014, 03:54

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