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# New Algebra Set!!!

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Intern
Joined: 13 May 2013
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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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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

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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Joined: 26 Jul 2012
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04 Jun 2013, 23:45
In question no.6)
how is m^2n^2=(mn)^2 ? Someone help me.
Math Expert
Joined: 02 Sep 2009
Posts: 60555

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05 Jun 2013, 00:40
In question no.6)
how is m^2n^2=(mn)^2 ? Someone help me.

$$(mn)^2 = (mn)*(mn)=m^2n^2$$.

Hope it's clear.
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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"?
Math Expert
Joined: 02 Sep 2009
Posts: 60555

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23 Jun 2013, 01:37
1
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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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 $$0+0-9=-9$$.

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).

Rgds,
TGC !!
Math Expert
Joined: 02 Sep 2009
Posts: 60555

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21 Jul 2013, 22:09
3
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 $$0+0-9=-9$$.

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).

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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Joined: 22 Jul 2013
Posts: 12
GMAT 1: 650 Q48 V31

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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.

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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 60555

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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.

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 ?

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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Schools: Sloan '18 (A)
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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.

Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power.
Math Expert
Joined: 02 Sep 2009
Posts: 60555

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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.

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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Location: India
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21 Jun 2014, 10:37
2
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.

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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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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Posts: 60555

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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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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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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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Joined: 28 Dec 2011
Posts: 4476

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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

Does all this make sense?
Mike
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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 $$0+0-9=-9$$.

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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Joined: 03 Jul 2014
Posts: 13

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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.

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!!!   [#permalink] 19 Nov 2014, 13:57

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