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# New Algebra Set!!!

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New Algebra Set!!! [#permalink]

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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

Kudos points for each correct solution!!!
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Re: New Algebra Set!!! [#permalink]

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21 Jun 2014, 02:56
Expert's post
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BOOKMARKED
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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Re: New Algebra Set!!! [#permalink]

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21 Jun 2014, 09:37
1
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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.

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 Algebra Set!!! [#permalink]

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10 Jul 2014, 01:56
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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Re: New Algebra Set!!! [#permalink]

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10 Jul 2014, 02:07
Expert's post
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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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Please explain .. not able to get it through !! [#permalink]

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03 Aug 2014, 07:11
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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05 Aug 2014, 10:40
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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Re: Please explain .. not able to get it through !! [#permalink]

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06 Aug 2014, 10:00
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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Re: New Algebra Set!!! [#permalink]

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15 Aug 2014, 23:44
Bunuel plz post a more clear soln for Q.2, I dont get it at all..Why is x^2 replaced as 3^3, and the end part..

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Re: New Algebra Set!!! [#permalink]

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28 Sep 2014, 12:53
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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Re: New Algebra Set!!! [#permalink]

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19 Nov 2014, 12:57
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

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New Algebra Set!!! [#permalink]

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20 Nov 2014, 06:36
hanschris5 wrote:
Bunuel wrote:
imhimanshu wrote:
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

No, it does NOT hold. You should plug into the original equation as I mention in my post you quote:

$$-2\neq{\sqrt[4]{(-2)^3+6(-2)^2}}$$.
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Re: New Algebra Set!!! [#permalink]

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20 Nov 2014, 08:45
These questions are not from GMATPrep Exam Pack 1. I know #6 is an exact copy of an official GMAT questions, but the others are not official GMAT questions.

Dabral

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Re: New Algebra Set!!! [#permalink]

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20 Nov 2014, 08:47
dabral wrote:
These questions are not from GMATPrep Exam Pack 1. I know #6 is an exact copy of an official GMAT questions, but the others are not official GMAT questions.

Dabral

Those are GMAT Club Tests question. Changed the tag. Thanks.
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Re: New Algebra Set!!! [#permalink]

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20 Nov 2014, 12:57
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.

Hello,
Can you pls tell me my error?
I took $$x^2$$ common and got $$x^2(x+6)$$ = $$x^4$$
then i divided everything by $$x^2$$.
So I only get two roots. x=-2 and x=3. I don't get x=0...

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Re: New Algebra Set!!! [#permalink]

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20 Nov 2014, 13:22
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 $$m=\frac{3}{n}$$.

Again, I did something basic wrong. I left 12 on the RHS, and took mn common to have mn(mn+1)=12.
mn=12 or mn=11....What is wrong with my concept?

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Re: New Algebra Set!!! [#permalink]

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20 Nov 2014, 22:30
usre123

By dividing both sides by $$x^2$$ you have eliminated the solution $$x=0$$.

For example, if we are given the condition $$x(x-1)=0$$, then there are two values of $$x$$ that satisfies this equation, $$x=0$$ and $$x=1$$. However, if we divide both sides by $$x$$ we end up with $$x-1=0$$, and left with only one solution. In general, on the GMAT never divide by a variable, unless you know that it is non-zero.

Cheers,
Dabral

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Re: New Algebra Set!!! [#permalink]

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20 Nov 2014, 22:34
"Again, I did something basic wrong. I left 12 on the RHS, and took mn common to have mn(mn+1)=12.
mn=12 or mn=11....What is wrong with my concept? "

You will find the mn=12 or mn=11 do not satisfy the above equation. We are looking for the same value of mn, this one is easy enough that many people can guess that mn=3 works, because (3)(3+1)=12, and the other solution is mn=4, which is a bit harder to spot. The better approach is to write it as a quadratic and then factor as shown in the solution.

Dabral

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Re: New Algebra Set!!! [#permalink]

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21 Nov 2014, 00:26
Got it, thank you!!

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Re: New Algebra Set!!! [#permalink]

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21 Nov 2014, 18:16
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, It is understood that even root of a negative expression is not possible. But here, for x= -2, the expression under root, x^3+6x^2 = -8 + 24 = 16 which is positive. Then why x = -2 invalid ?

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Re: New Algebra Set!!! [#permalink]

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21 Nov 2014, 19:45
^ -2 won't work if you put it on the LHS of the equation. LHS, x=-2? fourth root of what =-2? IT is not possible

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Re: New Algebra Set!!!   [#permalink] 21 Nov 2014, 19:45

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