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# New Algebra Set!!!

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New Algebra Set!!! [#permalink]  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]  26 Mar 2013, 01:07
Expert's post
Sinner1706 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,

Query here.

First of all we don't necessarily have to solve for all the three roots. Since, it is asking for a composite value i.e. the sum, we can directly obtain it.

We can rearrange to get:-

$$x^4-x^3-6x^2=0$$
Over here, by the system of poly equations Sum of roots = - ($$coeff ofx^3/coeff of x^4$$)
Similarly, Sum of roots taken two at a time = ($$coeff ofx^2/coeff of x^4$$)
Product of roots = "Zero" as there is no constant term.

Also, -2 satisfies the equation $$x^4-x^3-6x^2=0$$

So, shouldn't the answer be -1, with roots as 0,-2,3?

It's explained in the solution above: $$x$$ 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 $$x=3$$.
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Re: New Algebra Set!!! [#permalink]  26 Mar 2013, 04:03
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

Kudos points for each correct solution!!![/quote]

X^4 = X^3 + 6x^2.Solving this we get teh values of x as 0, 3 , -2.
But X cannot be negative as X is the positive fourth root of an expression(\sqrt{Expression}>= 0)
So X = 0 and 3
Sum of the values = 0+3 = 3
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Re: New Algebra Set!!! [#permalink]  26 Mar 2013, 04:12
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
Kudos points for each correct solution!!![/quote]

Since 3 is the root of the equation x^2 + ax + 15 = 0, therefore 3 must satisfy the equation x^2 + ax + 15 = 0 .
On putting x=3 in the equation x^2 + ax + 15 = 0 , we get a= -8
On putting the value of a in the original equation x^2 + ax - b = 0, we get x^2 - 8x - b = 0. As per question x^2 - 8x - b = 0 has equal roots so Discriminant(D) = b^2 -4ac = 0 = (-8)^2 - 4.1.(-b)= 0. From this we get the value of b as -16
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Re: New Algebra Set!!! [#permalink]  26 Mar 2013, 04:17
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. $$\frac{\sqrt{m^2+n^2}}{2}$$
Kudos points for each correct solution!!![/quote]

Adding the given two equations we get 2a^2 = (m+n)/2
Subtracting the given two equations we get 2b^2 = (m-n)/2

Multiplying above two equations we get 4a^2b^2 = (m^2 - n^2) => ab = $$\frac{\sqrt{m^2-n^2}}{2}$$
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Re: New Algebra Set!!! [#permalink]  26 Mar 2013, 04:29
4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?

A. -39
B. -9
C. 0
D. 9
E. 39
Kudos points for each correct solution!!![/quote]

Rearranging the given equation we get
-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 when x= 2 and y = -3, then the maximum value of the whole expression is 0 + 0 -9 = -9
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Re: New Algebra Set!!! [#permalink]  26 Mar 2013, 04:39
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
Kudos points for each correct solution!!![/quote]

On Re-arranging the given equation we get m = - x^2 - 2x +15

For m to be positive - x^2 - 2x +15 should be greater then 0. Solving we get (x + 5)(x - 3)<0. Since x is an integer we get the values of x as -4 , -3, -2, -1, 0, 1, 2 (total 7 values)

Given that X is an integer from -10 and 10, inclusive (21 values)

So, the probability is 7/21=1/3.

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Re: New Algebra Set!!! [#permalink]  26 Mar 2013, 04:44
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
Kudos points for each correct solution!!![/quote]

Rearranging we get m^2n^2 + mn - 12 = 0 => (mn + 4)(mn - 3)= 0 =>mn = -4 or mn = 3 => m= -4/n or 3/n
E. I and III only
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Re: New Algebra Set!!! [#permalink]  26 Mar 2013, 04:55
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
Kudos points for each correct solution!!![/quote]

Rearranging the equation we get x^4 - 29x^2 + 100 = 0 => (x^2 - 25)(x^ - 4) =0 => x= +5, -5, +2, -2
From these values of x we can get -50 (-5 * +5 * +2), +50(-5 * +5 * -2) but we can't get 25

B. II only
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Re: New Algebra Set!!! [#permalink]  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
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Re: New Algebra Set!!! [#permalink]  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
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Re: New Algebra Set!!! [#permalink]  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
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Re: New Algebra Set!!! [#permalink]  28 Mar 2013, 00:23
Expert's post
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.
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Re: New Algebra Set!!! [#permalink]  28 Mar 2013, 00:37
Expert's post
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
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Re: New Algebra Set!!! [#permalink]  28 Mar 2013, 00:49
Expert's post
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
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Re: New Algebra Set!!! [#permalink]  28 Mar 2013, 03:05
Expert's post
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
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Re: New Algebra Set!!! [#permalink]  28 Mar 2013, 03:10
Expert's post
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
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Re: New Algebra Set!!! [#permalink]  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
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Re: New Algebra Set!!! [#permalink]  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
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Re: New Algebra Set!!! [#permalink]  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 $$mn=3$$.

Solving and Factoring Quadratics:

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
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Re: New Algebra Set!!! [#permalink]  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
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Re: New Algebra Set!!!   [#permalink] 12 Apr 2013, 04:25

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