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# New Algebra Set!!!

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New Algebra Set!!!  [#permalink]

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Updated on: 18 Feb 2019, 05:31
53
198
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: https://gmatclub.com/forum/new-algebra- ... l#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: https://gmatclub.com/forum/new-algebra- ... l#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: https://gmatclub.com/forum/new-algebra- ... l#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: https://gmatclub.com/forum/new-algebra- ... l#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: https://gmatclub.com/forum/new-algebra- ... l#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: https://gmatclub.com/forum/new-algebra- ... l#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: https://gmatclub.com/forum/new-algebra- ... l#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: https://gmatclub.com/forum/new-algebra- ... l#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: https://gmatclub.com/forum/new-algebra- ... l#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: https://gmatclub.com/forum/new-algebra- ... l#p1200987

Kudos points for each correct solution!!!
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Originally posted by Bunuel on 18 Mar 2013, 07:56.
Last edited by Bunuel on 18 Feb 2019, 05:31, edited 1 time in total.
Updated.
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Re: New Algebra Set!!!  [#permalink]

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22 Mar 2013, 05:01
23
47
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.

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Re: New Algebra Set!!!  [#permalink]

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18 Mar 2013, 09:03
5
8. m^3 + 380 = 381m
m^3 + 380 = 380m + m
m^3 -m = 380m - 380
m(m-1)(m+1) = 380 (m-1)
m(m+1) = 380 ( -20 * -19; m is negative )
m = -20.

##### General Discussion
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Re: New Algebra Set!!!  [#permalink]

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22 Mar 2013, 06:30
18
41
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

Given $$m^3 + 380 = 380m+m$$.

Re-arrange: $$m^3-m= 380m-380$$.

$$m(m+1)(m-1)=380(m-1)$$. Since m is a negative integer, then $$m-1\neq{0}$$ and we can safely reduce by $$m-1$$ to get $$m(m+1)=380$$.

So, we have that 380 is the product of two consecutive negative integers: $$380=-20*(-19)$$, hence $$m=-20$$.

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Re: New Algebra Set!!!  [#permalink]

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22 Mar 2013, 05:16
10
28
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. $$\frac{\sqrt{m^2+n^2}}{2}$$

Sum the two equations: $$2a^2=m+n$$;
Subtract the two equations: $$2b^2=m-n$$;

Multiply: $$4a^2b^2=m^2-n^2$$;

Solve for $$ab$$: $$ab=\frac{\sqrt{m^2-n^2}}{2}$$

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Re: New Algebra Set!!!  [#permalink]

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22 Mar 2013, 06:44
10
13
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

$$f(x) = 2x - 1$$, hence $$f(n^2)=2n^2-1$$.
$$g(x) = x^2$$, hence $$g(n+12)=(n+12)^2=n^2+24n+144$$.

Since given that $$f(n^2)=g(n+12)$$, then $$2n^2-1=n^2+24n+144$$. Re-arranging gives $$n^2-24n-145=0$$.

Next, Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$:

$$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$.

Thus according to the above $$n_1*n_2=-145$$.

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Re: New Algebra Set!!!  [#permalink]

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22 Mar 2013, 06:08
7
35
Sorry, there was a typo in the stem .

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

Re-arrange the given equation: $$-x^2-2x+15=m$$.

Given that $$x$$ is an integer from -10 and 10, inclusive (21 values) we need to find the probability that $$-x^2-2x+15$$ is greater than zero, so the probability that $$-x^2-2x+15>0$$.

Factorize: $$(x+5)(3-x)>0$$. This equation holds true for $$-5<x<3$$.

Since x is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2.

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

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Re: New Algebra Set!!!  [#permalink]

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22 Mar 2013, 05:10
6
33
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

Since one of the roots of the equation $$x^2 + ax + 15 = 0$$ is 3, then substituting we'll get: $$3^2+3a+15=0$$. Solving for $$a$$ gives $$a=-8$$.

Substitute $$a=-8$$ in the first equation: $$x^2-8x-b=0$$.

Now, we know that it has equal roots thus its discriminant must equal to zero: $$d=(-8)^2+4b=0$$. Solving for $$b$$ gives $$b=-16$$.

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Re: New Algebra Set!!!  [#permalink]

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22 Mar 2013, 05:47
6
62
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 $$0+0-9=-9$$.

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Re: New Algebra Set!!!  [#permalink]

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18 Mar 2013, 08:22
4
3. a^2 + b^ 2 = m
a^2 - b^2 = n
Solving both the equations( adding them, and then subtracting them ):
2a^2 = m + n
2b^2 = m - n.
a = ((m+n)/2)^(1/2)
b = ((m-n)/2)^(1/2)

ab = ((m^2 - n^2)^(1/2))/2

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Re: New Algebra Set!!!  [#permalink]

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18 Mar 2013, 09:42
4
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

$$m^2n^2 + mn = 12$$
$$mn(mn+1) = 12$$
mn = 3
$$3(3+1)=12$$
OR
mn = -4
$$-4(-4+1)=12$$

$$m= 3/n$$
OR
$$m=-4/n$$
E
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Re: New Algebra Set!!!  [#permalink]

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22 Mar 2013, 06:20
4
8
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

Re-arrange and factor for x^2: $$(x^2-25)(x^2-4)=0$$.

So, we have that $$x=5$$, $$x=-5$$, $$x=2$$, or $$x=-2$$.

$$-50=5*(-5)*2$$;
$$50=5*(-5)*(-2)$$.

Only 25 is NOT a product of three possible values of x

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Re: New Algebra Set!!!  [#permalink]

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18 Mar 2013, 08:11
3
1
GyanOne wrote:
1. x^4 = x^3 + 6x^2
=> x^2 (x^2 - x - 6) = 0
Sum of roots of x^2 - x - 6 = 0 is 1 and the only other solution is x=0
=> Sum of all possible solutions = 1.

Option C

Some questions are tricky!!!
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Re: New Algebra Set!!!  [#permalink]

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18 Mar 2013, 09:33
3
1)$$x^4=x^3+6x^2$$
$$x^4-x^3-6x^2=0$$
$$x^2(x^2-x-6)=0$$
$$x^2=0 (1) x = 0$$
$$x^2-x-6=0 (2) x = 3 (3) x = -2$$
$$0+3-2=1$$
C

2)The equation x^2 + ax - b = 0 has equal roots
$$a^2+4b=0$$
one of the roots of the equation x^2 + ax + 15 = 0 is 3
$$(-a+-\sqrt{a^2-4*15})/2=3$$
$$+-\sqrt{a^2-60}=6+a$$
$$(\sqrt{a^2-60})^2=(6+a)^2$$
$$-96=12a$$
$$a=-8$$

$$a^2+4b=0$$
$$64+4b=0$$
$$b=-16$$
B

3)We can use some numbers
$$2^2+1^2=5=m 2^2-1^2=5=n$$
$$ab=2$$

$$\sqrt{m^2-n^2}/2 = \sqrt{25-9}/2 = 2 = ab$$
C

4) -3x^2 + 12x -2y^2 - 12y - 39 MAX
$$-3x^2 + 12x$$ has max in (2,12)
$$-2y^2 - 12y - 39$$ has max in (-6,-3)
-3x^2 + 12x -2y^2 - 12y - 39 MAX = 12 - 3 =9
D
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Re: New Algebra Set!!!  [#permalink]

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22 Mar 2013, 06:13
3
6
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 $$m=\frac{3}{n}$$.

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Re: New Algebra Set!!!  [#permalink]

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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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Re: New Algebra Set!!!  [#permalink]

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Updated on: 18 Mar 2013, 13:34
2
Edited

1. x^4 = x^3 + 6x^2
=> x^2 (x^2 - x - 6) = 0
The roots of x^2 - x - 6 are -2 and 3, but -2 cannot be the value of x. So 3 and 0 are the only possible roots.
=> Sum of all possible solutions = 3.

Option D
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Originally posted by GyanOne on 18 Mar 2013, 08:06.
Last edited by GyanOne on 18 Mar 2013, 13:34, edited 3 times in total.
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Re: New Algebra Set!!!  [#permalink]

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Updated on: 18 Mar 2013, 10:13
2
1
4. -3x^2 + 12x -2y^2 - 12y - 39 = 3(-x^2 + 4x - 13) + 2(-y^2-6y)

Now -x^2 +4x - 13 has its maximum value at x = -4/-2 = 2 and -y^2 - 6y has its maximum value at y=6/-2 = -3
Therefore max value of the expression
= 3(-4 + 8 -13) + 2 (-9+18) = -27 + 18 = -9

Should be B
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Originally posted by GyanOne on 18 Mar 2013, 08:23.
Last edited by GyanOne on 18 Mar 2013, 10:13, edited 1 time in total.
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Re: New Algebra Set!!!  [#permalink]

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18 Mar 2013, 08:40
2
6. M^2 * n^2 + mn = 12
mn( mn + 1 ) = 12
mn = 3 or mn = -4 ( 3 * 4 = 12; -4 * -3 = 12 )
So m = 3/n or m = -4/n

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Re: New Algebra Set!!!  [#permalink]

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18 Mar 2013, 11:55
2
@Bunuel, for Q5, shouldn't the original question also say that we are only looking for solutions where x is an integer?

For this to be satisfied, in x^2 + 2x -15 = -m, 4 - 4(m-15)>0 and 4-4(m-15) must be a perfect square
=> 4 (16-m) must be a perfect square and >0
=> 16-m must be a perfect square and >0
The only values that satisfy this for -10<=m<=10 are m=-9,0,7 of which only 7 is positive
=> probability = 1/3
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Re: New Algebra Set!!!   [#permalink] 18 Mar 2013, 11:55

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