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# New Algebra Set!!!

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Math Expert
Joined: 02 Sep 2009
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18 Mar 2013, 07:56
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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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Joined: 22 Oct 2017
Posts: 5

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06 Feb 2018, 10:21
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: $$3^2+3a+15=0$$. 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 $$b=-16$$.

Hi Bunuel,

hope you can help:

In the description it says "has equal roots, and one of the roots of the equation is ... 3".

I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" --> therefore i would have never thought about setting the determinant equal to zero.

Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero?

Thank you very much!

Best regards
Math Expert
Joined: 02 Sep 2009
Posts: 44388

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06 Feb 2018, 10:24
andr3 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: $$3^2+3a+15=0$$. 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 $$b=-16$$.

Hi Bunuel,

hope you can help:

In the description it says "has equal roots, and one of the roots of the equation is ... 3".

I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" --> therefore i would have never thought about setting the determinant equal to zero.

Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero?

Thank you very much!

Best regards

1. x^2 + ax - b = 0 has equal roots

2. One of the roots of the equation x^2 + ax + 15 = 0 is 3.

Those are two different equations.
_________________
Intern
Joined: 22 Oct 2017
Posts: 5

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06 Feb 2018, 10:42
Bunuel wrote:
andr3 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: $$3^2+3a+15=0$$. 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 $$b=-16$$.

Hi Bunuel,

hope you can help:

In the description it says "has equal roots, and one of the roots of the equation is ... 3".

I know that when an equation has only one solution, the determinant needs to be zero. BUT: in the description it says "and one of the roots is ... 3" --> therefore i would have never thought about setting the determinant equal to zero.

Could you please explain where the hint is to know that it's necessary to set the determinant equal to zero?

Thank you very much!

Best regards

1. x^2 + ax - b = 0 has equal roots

2. One of the roots of the equation x^2 + ax + 15 = 0 is 3.

Those are two different equations.

ok, than it is that language of "has equal roots", which I don't understand. could you clarify what "equal roots" means? is "equal roots" the definition of having only one solution?

thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 44388

### Show Tags

06 Feb 2018, 10:44
andr3 wrote:

ok, than it is that language of "has equal roots", which I don't understand. could you clarify what "equal roots" means? is "equal roots" the definition of having only one solution?

thanks!

Yes, equal roots means that there are two roots which are equal to each other, so one distinct root.
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Joined: 22 Oct 2017
Posts: 5

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06 Feb 2018, 10:56
Bunuel wrote:
andr3 wrote:

ok, than it is that language of "has equal roots", which I don't understand. could you clarify what "equal roots" means? is "equal roots" the definition of having only one solution?

thanks!

Yes, equal roots means that there are two roots which are equal to each other, so one distinct root.

ok, thank you very much!
Manager
Joined: 19 Aug 2016
Posts: 73

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17 Feb 2018, 05:19
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: $$3^2+3a+15=0$$. 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 $$b=-16$$.

Hi Bunuel

Can u pls elaborate on d=(-8)^2+4b?

Intern
Joined: 09 Feb 2018
Posts: 4

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22 Feb 2018, 22:47
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 $$0+0-9=-9$$.

can someone help me how the max value of the highlighted text is defined ?
Math Expert
Joined: 02 Sep 2009
Posts: 44388

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22 Feb 2018, 22:53
SaravanaPrabu090492 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 $$0+0-9=-9$$.

can someone help me how the max value of the highlighted text is defined ?

$$-3(x-2)^2= (-3)*(square \ of \ a \ number)= (negative)*(non-negative) = (non-positive)$$. So, this term is negative or 0.

The same for $$-2(y+3)^2$$.
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Joined: 09 Feb 2018
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22 Feb 2018, 23:42
Bunuel wrote:
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 $$n_1*n_2=-145$$.

Having a doubt with the above highlighted text. for the first, shouldn't the whole term [2x-1] be squared like the below one ? seeking solutions.
Math Expert
Joined: 02 Sep 2009
Posts: 44388

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23 Feb 2018, 00:15
1
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Expert's post
SaravanaPrabu090492 wrote:
Bunuel wrote:
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 $$n_1*n_2=-145$$.

Having a doubt with the above highlighted text. for the first, shouldn't the whole term [2x-1] be squared like the below one ? seeking solutions.

No.

$$f(x) = 2x - 1$$ means that we have a function which tells us what to do to get the value of the function: take a value for which you want to find the output of the function, multiply it by 2 and subtract 1.

So, to get $$f(n^2)$$ put n^2 instead of x: $$f(n^2)=2n^2-1$$.

Similarly for $$g(x) = x^2$$. To get $$g(n+12)$$ put n + 12 instead of x: $$g(n+12)=(n+12)^2=n^2+24n+144$$.
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