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# New Algebra Set!!!

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Math Expert
Joined: 02 Sep 2009
Posts: 56302

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25 Nov 2018, 02:21
shiv17 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 anybody help me with this solution. Why the maximum value of $$-3(x-2)^2$$ and $$-2(y+3)^2$$ is zero?

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Joined: 30 Oct 2018
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25 Nov 2018, 13:37
putting the value of x in eq -
3^2+3a+15=0
=> a= -8
Now putting a= -8 in x^2 + ax - b = 0
x^2−8x−b=0.

as roots are equal
=> b^2-4ac = 0
=> b = -16 (B)
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Plz Kudos! if my post helps....
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GMAT 1: 740 Q49 V41
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03 Jun 2019, 15:42
Bunuel wrote:
LaxAvenger wrote:
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 $$m=\frac{3}{n}$$.

Hello, how come m^2n^2 will become (mn)^2 ?

$$(mn)^2=(mn)(mn)=m^2*n^2$$.

Hi - I think this explains how mn^2 becomes m^2*n^2, but how does (m)^(2n^2) become (mn)^2? I put the parenthesis to show how I read the question, so perhaps my error and the error of the original questioner is in how it is written and not how it is solved. Thanks!
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Joined: 02 Sep 2009
Posts: 56302

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03 Jun 2019, 21:35
ghostman23 wrote:
Hi - I think this explains how mn^2 becomes m^2*n^2, but how does (m)^(2n^2) become (mn)^2? I put the parenthesis to show how I read the question, so perhaps my error and the error of the original questioner is in how it is written and not how it is solved. Thanks!

m^2n^2 can only mean $$m^2n^2$$, nothing else. If it were m^(2n^2), it would be written that way.
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08 Jun 2019, 10:04
Bunuel wrote:
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!!!

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

A quick way to solve this question, without having to find the value of a by substituting 3 in to equation 2 is to simply see that the second equation's constant value is 15. The roots of the second equation will have to equal 15 (think about what you do when factoring quadratic equations e.g. to find factors of ax^2 + bx + c you will look for two factors of AC that in some way add up to b).

So the other factor of the second equation is 5, because 3x5=15. So we know that a will be equal to 5 + 3 = 8. Once you know that you can rewrite the first equation as x^2 + 8x - b. Now remember that equation 1 has equal roots which means it follows the form of (a+b)^2 which expands to a^2 + 2ab + b^2.

Therefore a=2*1*4 (1 is the coefficient of x^2, and if a = 2 * first expression * second expression then a = 2*1*4). Hence b = 4^2= 16 (think about the (a+b)^2 form again). The negative sign is needed so the expression follows this form.

New Algebra Set!!!   [#permalink] 08 Jun 2019, 10:04

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