New Algebra Set!!! : GMAT Problem Solving (PS) - Page 6
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 10 Dec 2016, 18:36

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# New Algebra Set!!!

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 35951
Followers: 6866

Kudos [?]: 90120 [29] , given: 10418

### Show Tags

18 Mar 2013, 06:56
29
KUDOS
Expert's post
89
This post was
BOOKMARKED
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!!!
_________________
Intern
Joined: 15 Jan 2013
Posts: 39
Concentration: Finance, Operations
GPA: 4
Followers: 0

Kudos [?]: 27 [0], given: 6

### Show Tags

26 Mar 2013, 05:17
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

on re-arranging we get m^3 -381m = -380 => m(m^2 - 381)= 380
On checking the options and hit and trial, we get the values of m to be 1 and -20. But since m is negative so m = -20

B. -20
Intern
Joined: 15 Jan 2013
Posts: 39
Concentration: Finance, Operations
GPA: 4
Followers: 0

Kudos [?]: 27 [0], given: 6

### Show Tags

26 Mar 2013, 05:29
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

$$x=(\sqrt{5}-\sqrt{7})^2$$

x= 5 + 7 - 2. $$\sqrt{35}$$
$$\sqrt{35}$$ ~ 6

x = 5+7 - 2*6 = 12 - 12 = 0
Intern
Joined: 15 Jan 2013
Posts: 39
Concentration: Finance, Operations
GPA: 4
Followers: 0

Kudos [?]: 27 [0], given: 6

### Show Tags

26 Mar 2013, 05:34
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

f(n^2) = $$2.n^2 - 1$$
g(n + 12) = (n + 12)^2
f(n^2) = g(n + 12) => $$2.n^2 - 1$$ = (n + 12)^2 => 2.n^2 -1 = n^2 + 144 +24.n => n^2 -24n -145 = 0
From the above equation we get n = -5 and 29
So product of values of n = -5 * 29 = -145

A. -145
MBA Section Director
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 3723
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
Followers: 379

Kudos [?]: 2784 [0], given: 1961

### Show Tags

28 Mar 2013, 00:23
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

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
x=0 or 3 or -2. With x = -2 original equation does not hold true so possible values for X are 3 and 0 Hence their sum is 3
Choice D.

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

x^2 + ax - b = 0 that means a^2 - 4(-b) = 0 ------> a^2 + 4b = 0 -------------Equation I
x^2 + ax + 15 has 3 as a root, so 9 + 3a + 15 = 0 -------> 3a + 24 = 0 --------> a = -8
We will put the value of a in equation I -----------> a^2 + 4b = 0 --------> 64 + 4b = 0 ------> b = -16 ----> Choice B is the answer.
_________________
MBA Section Director
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 3723
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
Followers: 379

Kudos [?]: 2784 [0], given: 1961

### Show Tags

28 Mar 2013, 00:37
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

a^2 + b^2 = m
a^2 - b^2 = n
------------------------
2a^2 = m+n ----------> a^2 = (m+n)/2 --------> a = sq.root (m+n) / sq.root 2
similarly b = sq.root (m-n) / sq.root 2
So ab = (sq.root (m+n) / sq.root 2)(sq.root (m-n) / sq.root 2) -------> ab = sq.root(m^2 - n^2)/2---------------Using (a+b)(a-b) = a^2 - b^2
Choice C

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
_________________
MBA Section Director
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 3723
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
Followers: 379

Kudos [?]: 2784 [0], given: 1961

### Show Tags

28 Mar 2013, 00:49
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

there are 21 possible values of x so does m has 21 possible values
from x = -4 to x = 2, m gives negative value; Total 7 values
so the probability m being negative is 7/21----> 1/3 ----> choice B

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

m^2n^2 + mn = 12 ------> mn(mn+1) = 12 ----------> here mn and mn+1 are consecutive integers. So 12 is basically a product of 2 consecutive intergers
those integers must be 3 and 4 or -3 and -4
so possible values for mn are mn = 3 and mn+1 = 4 --------> m = (3/n)
mn = -4 and mn + 1 =-3 -------> m =-(4/n)
Choice E
_________________
MBA Section Director
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 3723
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
Followers: 379

Kudos [?]: 2784 [0], given: 1961

### Show Tags

28 Mar 2013, 03:05
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

x^4 - 29x^2 + 100 = 0 --------> let X^2 be y -------> y^2 - 29y + 100 = 0 ------> y^2 - 25y - 4y + 100 = 0 --------> (y-25)(y-4)=0
y=25 -----> x^2 = 25 -----> x= 5 or -5
y=4-------> x^2 = 4 ------> x = 2 or -2

Possible values of x are 5, -5, 2, and -2
Out of options 25 can never be a product of any 3 values of x, so Choice B is the answer

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

This was a bit complex one
m^3 - 381m = - 380 ------> m(m^2 - 381) = - 380
Since -380 is the product of m(m^2 - 381), any one of either m or (m^2 - 381) must be negative.
Since we are given that m is a negative integer, then m^2 - 381 must be positive
So m^2 - 381 > 0 -------> m^2 > 381 ---------> |m| > 19.5 ----approx.
so m > 19.5-------if m is positive and m < 19.5 ------ if m is negative
We know that m is negative so m must equal to -20 : choice B
_________________
MBA Section Director
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 3723
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
Followers: 379

Kudos [?]: 2784 [0], given: 1961

### Show Tags

28 Mar 2013, 03:10
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

x=(sq.root 5 - sq.root 7)^2 --------> x = 5 + 7 -2(sq.root 35) ---------using (a-b)^2 = a^2 + b^2 - 2ab
12 - 2(sq.root 35) ---------> assuming sq.root 35 = 6 ----------> 12-12 = 0
Choice A

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
_________________
Manager
Joined: 08 Dec 2012
Posts: 67
Location: United Kingdom
GMAT 1: 710 Q0 V0
WE: Engineering (Consulting)
Followers: 1

Kudos [?]: 181 [0], given: 31

### Show Tags

29 Mar 2013, 05:54
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.

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
MBA Section Director
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 3723
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
Followers: 379

Kudos [?]: 2784 [1] , given: 1961

### Show Tags

29 Mar 2013, 06:04
1
KUDOS
Expert's post
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.

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.
_________________
Manager
Joined: 12 Dec 2012
Posts: 230
GMAT 1: 540 Q36 V28
GMAT 2: 550 Q39 V27
GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)
Followers: 4

Kudos [?]: 76 [0], given: 181

### Show Tags

11 Apr 2013, 14:31
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 $$m=\frac{3}{n}$$.

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

My RC Recipe
http://gmatclub.com/forum/the-rc-recipe-149577.html

My Problem Takeaway Template
http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html

Math Expert
Joined: 02 Sep 2009
Posts: 35951
Followers: 6866

Kudos [?]: 90120 [1] , given: 10418

### Show Tags

12 Apr 2013, 01:22
1
KUDOS
Expert's post
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 $$m=\frac{3}{n}$$.

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 $$mn=3$$.

Hope it helps.
_________________
Manager
Joined: 12 Dec 2012
Posts: 230
GMAT 1: 540 Q36 V28
GMAT 2: 550 Q39 V27
GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)
Followers: 4

Kudos [?]: 76 [0], given: 181

### Show Tags

12 Apr 2013, 03:29
Bunuel wrote:

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

Hope it helps.

ohhhhhhhhhhhhhhhhhhhhh I thought that m is raised to the power 2n and the whole bracket raised to the power 2 ...that is what made me confused . Thanks a million , Bunuel
_________________

My RC Recipe
http://gmatclub.com/forum/the-rc-recipe-149577.html

My Problem Takeaway Template
http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html

Current Student
Joined: 04 Mar 2013
Posts: 69
Location: India
Concentration: Strategy, Operations
Schools: Booth '17 (M)
GMAT 1: 770 Q50 V44
GPA: 3.66
WE: Operations (Manufacturing)
Followers: 5

Kudos [?]: 52 [0], given: 27

### Show Tags

12 Apr 2013, 04:25
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.

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?

Anshuman
_________________

When you feel like giving up, remember why you held on for so long in the first place.

MBA Section Director
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 3723
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
Followers: 379

Kudos [?]: 2784 [0], given: 1961

### Show Tags

13 Apr 2013, 08:01
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
_________________
Senior Manager
Joined: 07 Sep 2010
Posts: 336
Followers: 6

Kudos [?]: 639 [0], given: 136

### Show Tags

22 May 2013, 21:19
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.

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
_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Math Expert
Joined: 02 Sep 2009
Posts: 35951
Followers: 6866

Kudos [?]: 90120 [0], given: 10418

### Show Tags

23 May 2013, 01:00
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.

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?
_________________
Intern
Joined: 13 May 2013
Posts: 12
Followers: 0

Kudos [?]: 14 [0], given: 27

### Show Tags

23 May 2013, 04:03
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

B
Intern
Joined: 13 May 2013
Posts: 12
Followers: 0

Kudos [?]: 14 [1] , given: 27

### Show Tags

23 May 2013, 04:13
1
KUDOS
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
Intern
Joined: 13 May 2013
Posts: 12
Followers: 0

Kudos [?]: 14 [1] , given: 27

### Show Tags

23 May 2013, 04:27
1
KUDOS
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
Re: New Algebra Set!!!   [#permalink] 23 May 2013, 04:27

Go to page   Previous    1   2   3   4   5   6   7   8   9    Next  [ 171 posts ]

Similar topics Replies Last post
Similar
Topics:
algebra 2 30 Jul 2011, 12:13
ALGEBRA 6 06 Jul 2011, 12:08
Algebra! 1 15 Jun 2011, 07:46
Algebra 1 01 May 2011, 12:51
Algebra 6 16 Sep 2010, 06:29
Display posts from previous: Sort by