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New Algebra Set!!!

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New post 23 May 2013, 05:13
1
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:

a^2 - b^2 = n => (a+b)(a-b)=n=> (a+b)^2*(a-b)^2=n^2 => (a^2 + b^2+2ab)*(a^2 + b^2-2ab)=n^2 => (m+2ab)*(m-2ab)=n^2 => m^2-2ab^2=n^2
=>2ab^2=m^2-n^2 => 2ab=\sqrt{\(m^2-n^2\)} =>ab=\sqrt{\(m^2-n^2\)}/2

Ans:
C
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New post 23 May 2013, 05:27
1
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
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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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New post 05 Jun 2013, 00:40
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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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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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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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New post 21 Jul 2013, 22:09
1
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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New post 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
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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 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.
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Re: New Algebra Set!!!  [#permalink]

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New post 21 Jun 2014, 03:56
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.
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Re: New Algebra Set!!!  [#permalink]

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New post 21 Jun 2014, 10:37
1
1
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.


From GMATCLUB math book:

General rules:
• \(\sqrt{x}\sqrt{y}=\sqrt{xy}\) and \(\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}\).

• \((\sqrt{x})^n=\sqrt{x^n}\)

• \(x^{\frac{1}{n}}=\sqrt[n]{x}\)

• \(x^{\frac{n}{m}}=\sqrt[m]{x^n}\)

• \({\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}\)

• \(\sqrt{x^2}=|x|\), when \(x\leq{0}\), then \(\sqrt{x^2}=-x\) and when \(x\geq{0}\), then \(\sqrt{x^2}=x\)

• 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 a positive value on the GMAT.

• Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).


More on this can be found here:
math-number-theory-88376.html#p666609
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New post 10 Jul 2014, 02:56
Bunuel wrote:
SOLUTIONs:

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 \(x=3\).




Can sumone plz explain the part in the quote? I dont quite get it :( I read WoundedTiger's post (one above) as well i still do not get it :( a lil help in layman terms plz


If the entire term under the 4th sq root , say is 16
then my x can be a -2 ryt?
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New post 10 Jul 2014, 03:07
janxavier wrote:
Bunuel wrote:
SOLUTIONs:

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 \(x=3\).




Can sumone plz explain the part in the quote? I dont quite get it :( I read WoundedTiger's post (one above) as well i still do not get it :( a lil help in layman terms plz


If the entire term under the 4th sq root , say is 16
then my x can be a -2 ryt?


\(\sqrt[4]{16}=2\), 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 03 Aug 2014, 08:11
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) −116 e) −164
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Re: New Algebra Set!!!  [#permalink]

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New post 05 Aug 2014, 11:40
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

[m]-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9




How do you know to break the 39 up into 12, 18, and 9?
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New post 06 Aug 2014, 11:00
1
vik09 wrote:
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) −116 e) −164

Dear vik09,
This is a tricky question, and I am happy to help. :-)

First of all, you may find these blogs helpful:
http://magoosh.com/gmat/2012/foil-on-th ... expanding/
http://magoosh.com/gmat/2012/algebra-on ... to-factor/

Let's start with x^2 + ax + 15 = 0. If one root is 3, then one factor must be (x − 3). Let's say the other root is [b]k[/b], and the other factor is (x − k). If we multiply (x - 3)(x - k), the constant term will be (3k), which has to equal 15. Therefore, k = 5, and

(x − 3)(x − 5) = x^2 − 8x + 15

Therefore, a = 8.

Now, we have x^2 + 8x − b = 0, which has equal roots. If a quadratic has equal roots, it must be the square of either a sum or a difference:
Square of a sum: (a + b)^2 = a^2 + 2ab + b^2
Square of a difference: (a − b)^2 = a^2 − 2ab + b^2
See:
http://magoosh.com/gmat/2013/three-alge ... -the-gmat/

Look at the square of a sum formula. We replace a = x. In order for the middle term to equal 8x, we must have b = 4, and the full equation would be

(x + 4)^2 = x^2 + 8x + 16

Setting the final terms equal, we have
− b = 16
b = −16
Answer = (B)

Does all this make sense?
Mike :-)
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Re: New Algebra Set!!!  [#permalink]

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New post 28 Sep 2014, 13:53
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.

Is there another solution in which we don't need to manipulate the number so much?
How do you know when there is a shortcut like used here and when there isn't?
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Re: New Algebra Set!!!  [#permalink]

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New post 19 Nov 2014, 13:57
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?





well u can plug in -2 and still it holds.

x^4 = (-2)^4 = 16 on the other side of equal to ie x^3+6x^2 we can write (-2)^3+6(-2)^2 equals -8+24 = 16 .....so holds :) ....please correct me . Also BTW in solving quadratic equation such as x^2-x+6=0 we do consider both the roots -ve and +ve , until its stated not to ... thanks
Re: New Algebra Set!!! &nbs [#permalink] 19 Nov 2014, 13:57

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