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Inequalities Made Easy!

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Intern
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Joined: 22 Apr 2019
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Re: Inequalities Made Easy!  [#permalink]

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New post 08 Jun 2019, 07:06
Bunuel wrote:

Notice that \((x+a)(x+b)(x+c)>0\) can be written as \((x-(-a))(x-(-b))(x-(-c))>0\). So, if the "roots" of (x-a)(x-b)(x-c)>0 are a, b, and c, for \((x-(-a))(x-(-b))(x-(-c))>0\) they would be -a, -b, and -c.

For example, for (x + 1)(x + 2) > 0, you put on the line -1 and -2.

Hope it helps.


Ah perfect! Thanks!
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Re: Inequalities Made Easy!  [#permalink]

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New post 23 Jul 2019, 18:18
Hi Karishma & Bunuel,

First of all, let me thank you to put together such an amazing explanation of Inequalities and strategies. It helped me immensely.

I have a small doubt in Step 2 mentioned in the very first post "Inequalities with Multiple Factors" of this thread:

"Starting from the rightmost section, mark the sections with alternate positive and negative signs. The inequality will be positive in the sections where you have the positive signs and it will be negative in the sections where you have the negative signs"

My understanding is that we have to check one value from each section to understand if the equation results in a Positive or Negative value in order to assign the sign to each of the section. is this correct? or we can blindly assign Positive and Negative sign to alternate sections respectively by starting with assigning Positive sign to the rightmost section?
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Re: Inequalities Made Easy!  [#permalink]

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New post 23 Jul 2019, 18:22
shishirdixit wrote:
Hi Karishma & Bunuel,

First of all, let me thank you to put together such an amazing explanation of Inequalities and strategies. It helped me immensely.

I have a small doubt in Step 2 mentioned in the very first post "Inequalities with Multiple Factors" of this thread:

"Starting from the rightmost section, mark the sections with alternate positive and negative signs. The inequality will be positive in the sections where you have the positive signs and it will be negative in the sections where you have the negative signs"

My understanding is that we have to check one value from each section to understand if the equation results in a Positive or Negative value in order to assign the sign to each of the section. is this correct? or we can blindly assign Positive and Negative sign to alternate sections respectively by starting with assigning Positive sign to the rightmost section?


You can blindly assign assuming that it's set up correctly.
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Re: Inequalities Made Easy!  [#permalink]

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New post 05 Aug 2019, 16:01
Hello Bunuel,

How did you factor this?

"Given: −2x3+17x2–30x>0−2x3+17x2–30x>0

x(−2x2+17x–30)>0x(−2x2+17x–30)>0 (taking x common)

x(2x–5)(6–x)>0x(2x–5)(6–x)>0 (factoring the quadratic)"

Thanks
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Re: Inequalities Made Easy!   [#permalink] 05 Aug 2019, 16:01

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