New Algebra Set!!! : Problem Solving (PS)

WoundedTiger

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

19 Mar 2013, 17:39

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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

Sol: Adding the 2 equations we get a= \sqrt{(m+n)/2}
Subtracting Eqn 2 from 1, we get b= \sqrt{m-n/2}

Ans Option C

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

19 Mar 2013, 17:40

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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

Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get

3^2 +3a+15=0----> Solving for a we get a=-8.

Putting the value if a in Eqn 1 we get
x^2 -8x-b=0
We know for Eqn ax^2 +bx+c=-0
Sum of roots is given by -b/a
Product of roots is c/a

Therefore from Eqn1 (after subsitution of value of a) we get
Sum of roots = -8 and roots are equal ie. -4 and -4
Product of roots will be 16. Therefore Option B

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

20 Mar 2013, 02:32

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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.
B.
C.
D.
E.

Soln: Take equation 2, square on both sides, substitute for a^2 + b^2 with m, Ans (C)

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

Soln: Used a 'gestimation' method. Since the unit digit of the resulting values on both sides has to be same, the only number from the options that will fit this condition is -20. Ans (B)

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

20 Mar 2013, 08:58

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1. If x = (x^3 + 6x^2)^1/4, then the sum of all possible solutions for x is:

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

My take ==> C

Approach:

x = (x^3 + 6x^2)^1/4
x^4 = x^3 + 6x^2
x^4 - x^3 - 6x^2 = 0
x^2(x^2 - x - 6) = 0
x^2 * (x-3) * (x+2) = 0

So Sum of all the possible solution of x = 1

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

21 Mar 2013, 22:11

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7 - B

x^4=29x^2-100
x^4-29x^2+100=0
which is equal to

(x-2)(x+2)(x-5)(x+5) = 0

roots are 2,-2,5,-5

thus B is the right answer ==> 25 can't be the result of the multiplication of any three roots.

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

23 Mar 2013, 03:54

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Hi Bunuel,
Thanks for the challenging ones.

Quote:

Given \(m^3 + 380 = 380m+m\).

I was able to backsolve this question, as I couldn't able to figure out the above step. Can you please post your thinking pattern behind this question. I mean, how did you think about breaking the term 381m into 380m + m --->(A)
Any particular indicators in the question that make you do so.
I'm sure during exam time, I will not be able to do so.

Quote:

Re-arrange: \(m^3-m= 380m-380\).

\(m(m+1)(m-1)=380(m-1)\). Since m is a negative integer, then \(m-1\neq{0}\) and we can safely reduce by \(m-1\) to get \(m(m+1)=380\).

So, we have that 380 is the product of two consecutive negative integers: \(380=-20*(-19)\), hence \(m=-20\).

Answer: B.

I have read somewhere that cancelling the like terms in GMAT is dangerous. I can sense that in your post as well( Red colored part). Request you to please enlighten me with this concept as why did you need to make extra caution while cancelling (m-1) terms, instead of cancelling it without given a thought.

Thanks
H

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Bunuel

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

24 Mar 2013, 03:44

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imhimanshu wrote:

Quote:

Re-arrange: \(m^3-m= 380m-380\).

\(m(m+1)(m-1)=380(m-1)\). Since m is a negative integer, then \(m-1\neq{0}\) and we can safely reduce by \(m-1\) to get \(m(m+1)=380\).

So, we have that 380 is the product of two consecutive negative integers: \(380=-20*(-19)\), hence \(m=-20\).

Answer: B.

I have read somewhere that cancelling the like terms in GMAT is dangerous. I can sense that in your post as well( Red colored part). Request you to please enlighten me with this concept as why did you need to make extra caution while cancelling (m-1) terms, instead of cancelling it without given a thought.

Thanks
H

Consider equation xy=x. Here we cannot reduce by x, because x could be zero and we cannot divide by zero. What we can do is: xy=x --> xy-x=0 --> x(y-1)=0 --> x=0 or y=1.

Now, in the question we have \(m(m+1)(m-1)=380(m-1)\) and we know that m is a negative number, thus m-1 does not equal to 0, which means that we can safely reduce by m-1.

Hope it's clear.

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

11 Apr 2013, 15:31

Bunuel wrote:

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

Re-arrange: \((mn)^2 + mn - 12=0\).

Factorize for mn: \((mn+4)(mn-3)=0\). Thus \(mn=-4\) or \(mn=3\).

So, we have that \(m=-\frac{4}{n}\) or \(m=\frac{3}{n}\).

Answer: E.

I still cannot get how did you rearrange the given expression, bunuel . Could you please elaborate more ? Thanks in advance

L

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

12 Apr 2013, 02:22

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TheNona wrote:

Bunuel wrote:

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

Re-arrange: \((mn)^2 + mn - 12=0\).

Factorize for mn: \((mn+4)(mn-3)=0\). Thus \(mn=-4\) or \(mn=3\).

So, we have that \(m=-\frac{4}{n}\) or \(m=\frac{3}{n}\).

Answer: E.

I still cannot get how did you rearrange the given expression, bunuel . Could you please elaborate more ? Thanks in advance

\(m^2n^2 + mn = 12\) --> \(m^2n^2 + mn -12=0\) --> \((mn)^2 + mn - 12=0\) --> say mn=x, so we have \(x^2+x-12=0\) --> \((x+4)(x-3)=0\) --> \(x=-4\) or \(x=3\) --> \(mn=-4\) or \(mn=3\).

Solving and Factoring Quadratics:
https://www.purplemath.com/modules/solvquad.htm
https://www.purplemath.com/modules/factquad.htm

Hope it helps.

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

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"?

L

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

23 Jun 2013, 01:37

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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]

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 !!

Rgds,
TGC !!

L

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

21 Jul 2013, 22:09

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targetgmatchotu wrote:

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 !!

Rgds,
TGC !!

I completed the squares for -3x^2 + 12x - ... and for -2y^2 - 12y-... So, I asked myself what do I need there in order to have (a+b)^2.

Hope it's clear.

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

23 Oct 2013, 23:54

Bunuel wrote:

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?

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

L

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

24 Oct 2013, 01:01

Expert Reply

NeetiGupta 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]

21 Jun 2014, 01:35

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.

Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power.

L

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

21 Jun 2014, 03:56

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Ergenekon 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.

Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power.

x cannot be negative cannot be negative as it equals to the even (4th) root of some expression.

gmatclubot

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

21 Jun 2014, 03:56

GMAT Problem Solving (PS) Questions

New Algebra Set!!!