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.
thanks in advance.

\(-3x^2 + 12x -2y^2 - 12y - 39\) -->\(x(12-3x) - y(2y+12)-39\)

now x(12-3x) will be highest when x is 1.

-y(2y+12) will be higest when y = -1.
so the answer becomes 9+10-39 = -19..
where am I going wrong?
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unable to understand this line. Can someone please explain?
many thanks
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Check this: Solving Quadratic Inequalities: Graphic Approach

Fro more: check

Inequalities

• Theory
  Inequalities Made Easy!
  Inequalities: Tips and hints
  Arithmetic with Inequalities
  Wavy Line Method Application - Complex Algebraic Inequalities
  Solving Quadratic Inequalities: Graphic Approach
  Inequalities - Quadratic Inequalities
  Graphic approach to problems with inequalities
  Inequalities trick
  INEQUATIONS(Inequalities) Part 1
  INEQUATIONS(Inequalities) Part 2
  
  • Questions
    Inequality and absolute value questions from my collection
    DS Questions
    PS Questions

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Can anybody help me with this solution. Why the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero?

Please read the whole thread before posting: https://gmatclub.com/forum/new-algebra- ... l#p2020076
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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)

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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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.

Answer is thus B -16

Hi Bunuel,

Why can I not substract x^2 + ax + 15 = 0 from x^2 + ax - b = 0 and get b= -15?

Thanks!
Lionila

In this question -

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.


You didn't take the negative values - why are we considering negative values here for 4th roots?

Thanks Bunuel.
Apologies for that! I kept getting confused!
Even roots (sq root and 4th root) cannot have a -ve value but x^2=16 and x^4=16 can have (+2 or-2) as the solutions. I got it Thanks again!

Hi Bunuel, Can you please explain how did you rearrange the equation into a quadratic equation? Thanks

Hi Bunuel,
In first question why x cannot be negative . i know the square root is always positive . but when i put the value of x =-2 it staisfy both the side of equation which is creating a confusion for me.

How did you got maximum for x and y please explain ?

Re-arrange: How did u got this ? \((mn)^2 + mn - 12=0\).

Subtract 12 from both sides of \(m^2n^2 + mn = 12\) to get \((mn)^2 + mn - 12=0\).
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In the answer to question 6, Bunuel rearranges this expression "m^2n^2" to this (mn)^2. I would be appreciative if someone could explain the what and how.

In question 8, you, Bunuel, did this:

Given m3+380=380m+m.

Re-arrange: m3−m=380m−380m3−m=380m−380.

This lead you straight to the answer.

When I see an equation like the one above, I know that I have to rearrange the equation; however, given the countless ways you can rearrange an equation like that, I dont know which way to choose.

How do you decide which way to rearrange equations? On a question like 8, what is your thought process? How important do you think considered rearranging is?

Thank you so much for your help. You are great!