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# New Algebra Set!!!

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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

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23 May 2013, 04:42
SathyaNIT wrote:
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

The correct answer is B, not E: new-algebra-set-149349-60.html#p1200975

Notice that we are asked "which of the following is NOT a product of three possible values of x".

Hope it helps.
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25 May 2013, 22:27
Bunuel wrote:
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 $$ab$$: $$ab=\frac{\sqrt{m^2-n^2}}{2}$$

Answer: C.

Another method is picking numbers:
let a=2
b=3
m=a^2+b^2 ie 2*2+3*3= 13

Similarly,

n= a^2-b^2 ie 2*2-3*3= -5

Now we have to find a*b ie 2*3=6

plug the values of m=13 and n=-5

Option C gives 6 which we are looking for....

Thus Answer is C..
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04 Jun 2013, 22:45
In question no.6)
how is m^2n^2=(mn)^2 ? Someone help me.
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04 Jun 2013, 23:40
pradeepkamp wrote:
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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05 Jun 2013, 06:08
thank you Bunuel,,, I read the question differently ( m raised to 2n raised to 2)
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22 Jun 2013, 23:35
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"?
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23 Jun 2013, 00:37
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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08 Jul 2013, 23:40
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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21 Jul 2013, 21:05
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 $$0+0-9=-9$$.

Answer: B.

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).

Your reply is appreciated !!

Rgds,
TGC !!
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21 Jul 2013, 21:09
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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 $$0+0-9=-9$$.

Answer: B.

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).

Your reply is appreciated !!

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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10 Oct 2013, 16:57
Bunuel wrote:
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: $$3^3+3a+15=0$$. 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 $$b=-16$$.

Answer: B.

Can you explain what a root of an equation is, as well as a discriminant, I didn't get anything done for this one
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11 Oct 2013, 01:49
AccipiterQ wrote:
Bunuel wrote:
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: $$3^3+3a+15=0$$. 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 $$b=-16$$.

Answer: B.

Can you explain what a root of an equation is, as well as a discriminant, I didn't get anything done for this one

Check here: http://www.purplemath.com/modules/solvquad4.htm

Hope it helps.
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23 Oct 2013, 22:54
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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.

Answer: D.

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 ?
Please explain
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24 Oct 2013, 00:01
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.

Answer: D.

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 ?
Please explain

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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05 Apr 2014, 02:42
Q1.
Option D.
Raising both LHS and RHS to fourth power,and making RHS=0 we get
$$x^4-x^3-6x^2=0$$
Taking $$x^2$$ common and factorizing,
$$x^2(x+2)(x-3)=0$$
Three possible values of $$x$$ are $$-2,3$$ and $$0$$
But $$x$$ can't be $$-ve$$.
So sum of possible values=$$3$$
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05 Apr 2014, 02:47
Q2
Option B.
From second equation,calculate value of $$a$$ by putting $$x=3$$
$$a=-8$$
For roots to be equal,$$D=0$$
$$a^2+4b=0$$
$$b=-16$$

Last edited by AKG1593 on 05 Apr 2014, 02:54, edited 1 time in total.
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05 Apr 2014, 02:54
Q3.
Option C.
$$a^2+b^2=m$$
$$a^2-b^2=n$$
$$=>a^2=m+n/2$$
Or $$a=\sqrt{(m+n)/2}$$

Similarly $$b=\sqrt{(m-n)/2}$$
Multiplying $$a*b$$$$=\sqrt{(m^2-n^2)/4}$$
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20 May 2014, 01:02
Bunuel wrote:
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: $$3^3+3a+15=0$$. 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 $$b=-16$$.

Answer: B.

Why can't we simply deduct the first equation from the second so that, x^2 ans ax gets eliminated and b remains with a value of -15?
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20 May 2014, 01:25
farzana87 wrote:
Bunuel wrote:
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: $$3^3+3a+15=0$$. 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 $$b=-16$$.

Answer: B.

Why can't we simply deduct the first equation from the second so that, x^2 ans ax gets eliminated and b remains with a value of -15?

Well, because these equations are not the same: x^2-8x+16=0 and x^2-8x+15=0.
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21 Jun 2014, 00:35
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.

Answer: D.

Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power.
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# New Algebra Set!!!

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New Algebra Set!!! [#permalink]

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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

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Re: New Algebra Set!!! [#permalink]

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23 May 2013, 04:42
SathyaNIT wrote:
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

The correct answer is B, not E: new-algebra-set-149349-60.html#p1200975

Notice that we are asked "which of the following is NOT a product of three possible values of x".

Hope it helps.
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Re: New Algebra Set!!! [#permalink]

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25 May 2013, 22:27
Bunuel wrote:
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 $$ab$$: $$ab=\frac{\sqrt{m^2-n^2}}{2}$$

Answer: C.

Another method is picking numbers:
let a=2
b=3
m=a^2+b^2 ie 2*2+3*3= 13

Similarly,

n= a^2-b^2 ie 2*2-3*3= -5

Now we have to find a*b ie 2*3=6

plug the values of m=13 and n=-5

Option C gives 6 which we are looking for....

Thus Answer is C..
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Re: New Algebra Set!!! [#permalink]

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04 Jun 2013, 22:45
In question no.6)
how is m^2n^2=(mn)^2 ? Someone help me.
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Re: New Algebra Set!!! [#permalink]

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04 Jun 2013, 23:40
pradeepkamp wrote:
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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Re: New Algebra Set!!! [#permalink]

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05 Jun 2013, 06:08
thank you Bunuel,,, I read the question differently ( m raised to 2n raised to 2)
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Re: New Algebra Set!!! [#permalink]

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22 Jun 2013, 23:35
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"?
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Re: New Algebra Set!!! [#permalink]

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23 Jun 2013, 00:37
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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Re: New Algebra Set!!! [#permalink]

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08 Jul 2013, 23:40
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: New Algebra Set!!! [#permalink]

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21 Jul 2013, 21:05
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 $$0+0-9=-9$$.

Answer: B.

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).

Your reply is appreciated !!

Rgds,
TGC !!
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Re: New Algebra Set!!! [#permalink]

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

Answer: B.

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).

Your reply is appreciated !!

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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10 Oct 2013, 16:57
Bunuel wrote:
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: $$3^3+3a+15=0$$. 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 $$b=-16$$.

Answer: B.

Can you explain what a root of an equation is, as well as a discriminant, I didn't get anything done for this one
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Re: New Algebra Set!!! [#permalink]

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11 Oct 2013, 01:49
AccipiterQ wrote:
Bunuel wrote:
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: $$3^3+3a+15=0$$. 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 $$b=-16$$.

Answer: B.

Can you explain what a root of an equation is, as well as a discriminant, I didn't get anything done for this one

Check here: http://www.purplemath.com/modules/solvquad4.htm

Hope it helps.
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Re: New Algebra Set!!! [#permalink]

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23 Oct 2013, 22:54
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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.

Answer: D.

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 ?
Please explain
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Re: New Algebra Set!!! [#permalink]

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24 Oct 2013, 00:01
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.

Answer: D.

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 ?
Please explain

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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Re: New Algebra Set!!! [#permalink]

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05 Apr 2014, 02:42
Q1.
Option D.
Raising both LHS and RHS to fourth power,and making RHS=0 we get
$$x^4-x^3-6x^2=0$$
Taking $$x^2$$ common and factorizing,
$$x^2(x+2)(x-3)=0$$
Three possible values of $$x$$ are $$-2,3$$ and $$0$$
But $$x$$ can't be $$-ve$$.
So sum of possible values=$$3$$
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Re: New Algebra Set!!! [#permalink]

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05 Apr 2014, 02:47
Q2
Option B.
From second equation,calculate value of $$a$$ by putting $$x=3$$
$$a=-8$$
For roots to be equal,$$D=0$$
$$a^2+4b=0$$
$$b=-16$$

Last edited by AKG1593 on 05 Apr 2014, 02:54, edited 1 time in total.
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Re: New Algebra Set!!! [#permalink]

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05 Apr 2014, 02:54
Q3.
Option C.
$$a^2+b^2=m$$
$$a^2-b^2=n$$
$$=>a^2=m+n/2$$
Or $$a=\sqrt{(m+n)/2}$$

Similarly $$b=\sqrt{(m-n)/2}$$
Multiplying $$a*b$$$$=\sqrt{(m^2-n^2)/4}$$
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Re: New Algebra Set!!! [#permalink]

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20 May 2014, 01:02
Bunuel wrote:
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: $$3^3+3a+15=0$$. 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 $$b=-16$$.

Answer: B.

Why can't we simply deduct the first equation from the second so that, x^2 ans ax gets eliminated and b remains with a value of -15?
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New Algebra Set!!! [#permalink]

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20 May 2014, 01:25
farzana87 wrote:
Bunuel wrote:
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: $$3^3+3a+15=0$$. 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 $$b=-16$$.

Answer: B.

Why can't we simply deduct the first equation from the second so that, x^2 ans ax gets eliminated and b remains with a value of -15?

Well, because these equations are not the same: x^2-8x+16=0 and x^2-8x+15=0.
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Re: New Algebra Set!!! [#permalink]

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21 Jun 2014, 00:35
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.

Answer: D.

Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power.
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Re: New Algebra Set!!!   [#permalink] 21 Jun 2014, 00:35

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