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

Sol: The given eqn can be written as mn (mn + 1) = 12
Now mn, mn+ will have to be consecutive Integers and possible combinations are

mn=3 and mn+ 1= 4 ------> possible value of m will be 3/n
or mn+1 = -4 and mn= -3 -------. possible value of m will be -3/n

Ans should be III only i.e Option C
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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

Sol: Consider eqn 2 first. Since 3 is the root of the eqn we get

3^2 +3a+15=0----> Solving for a we get a=-8.

Putting the value if a in Eqn 1 we get
x^2 -8x-b=0
We know for Eqn ax^2 +bx+c=-0
Sum of roots is given by -b/a
Product of roots is c/a

Therefore from Eqn1 (after subsitution of value of a) we get
Sum of roots = -8 and roots are equal ie. -4 and -4
Product of roots will be 16. Therefore Option B
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[quote="ConnectTheDots"]7. If\(x^4 = 29x^2 - 100\), then which of the following is NOT a product of two possible values of x?

I. -50
II. 25
III. 100

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

\(x^4 = 29x^2 - 100\)
\(=> x^4-29x^2+100 = 0\)
\(=> x^4 - 25x^2 - 4x^2 + 100=0\)
\(=> x^2(x^2-25) - 4(x^2-25) = 0\)
\(=>(x^2 - 4)(x^2-25) = 0\)
=> x = +2,-2,+5,-5

None of -50, 25 or 100 is possible with product of two possible values of x.
what I am missing ...


---> you missed the q ,it is
7. If x^4 = 29x^2 - 100, then which of the following is NOT a product of three possible values of x
and answer is 25

I agree. Proper answer to 5 as written is 10/21

If x is not explicitly constrained or if x is not constrained by the equation given, then we have to respect all real values of x, integer or not. Since x is not constrained to integers only, we have to respect the possibility that x is not an integer.

Since -10 <= m <= 10 and the equation is x^2 + 2x - 15 = -m, we will always have x^2 + 2x + c = 0 where -25 <= c <= -5. Using quadratic formula (just to prove it, though you won't need to know it for the GMAT), we need the calculate the discriminant, which is b^2 - 4ac, to determine whether or not there are real values for x. Here we get 2^2 - 4*1*c. Since c is always negative, we will always have a positive number for the discriminant. This means that we will always have a positive number underneath the square root of the quadratic formula and, therefore, 2 real outcomes for x.

There are 21 integer values of m (10 positive, 10 negative, and 0) and 10 of them are positive, so 10/21.

Good question though if we constrain x to only integer values.

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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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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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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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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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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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

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

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

Answer
B

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.
_________________

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"?

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 !!

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.
_________________

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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