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 Post subject: what values of x that satisfy I5-2xI<=(less than equal to) 7
PostPosted: Thu, 15 Sep 2011 00:12:42 UTC 
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How do we solve inequalities such as the one mentioned above where there is absolute value?


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 Post subject: Re: what values of x that satisfy I5-2xI<=(less than equal t
PostPosted: Thu, 15 Sep 2011 00:25:01 UTC 
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You can rewrite it as -7 ≤ 5 - 2x ≤ 7.
-12 ≤ -2x ≤ 2, 12 ≥ 2x ≥ -2, 6 ≥ x ≥ -1.


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 Post subject: Re: what values of x that satisfy I5-2xI<=(less than equal t
PostPosted: Thu, 15 Sep 2011 05:38:41 UTC 
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mathematic wrote:
You can rewrite it as -7 ≤ 5 - 2x ≤ 7.
-12 ≤ -2x ≤ 2, 12 ≥ 2x ≥ -2, 6 ≥ x ≥ -1.


Square both sides.

4x^2-20x+25\le 49\iff 4x^2-20x-24\le 0, divide that annoying 4 out:

x^2-5x-6\le 0

factor:

(x-6)(x+1)\le 0

From here it's easy.

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 Post subject: Re: what values of x that satisfy I5-2xI<=(less than equal t
PostPosted: Thu, 15 Sep 2011 22:57:19 UTC 
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Shadow wrote:
mathematic wrote:
You can rewrite it as -7 ≤ 5 - 2x ≤ 7.
-12 ≤ -2x ≤ 2, 12 ≥ 2x ≥ -2, 6 ≥ x ≥ -1.


Square both sides.

4x^2-20x+25\le 49\iff 4x^2-20x-24\le 0, divide that annoying 4 out:

x^2-5x-6\le 0

factor:

(x-6)(x+1)\le 0

From here it's easy.

It's a matter of opinion, but I think my way is easier.


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 Post subject: Re: what values of x that satisfy I5-2xI<=(less than equal t
PostPosted: Thu, 15 Sep 2011 23:57:36 UTC 
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mathematic wrote:
Shadow wrote:
mathematic wrote:
You can rewrite it as -7 ≤ 5 - 2x ≤ 7.
-12 ≤ -2x ≤ 2, 12 ≥ 2x ≥ -2, 6 ≥ x ≥ -1.


Square both sides.

4x^2-20x+25\le 49\iff 4x^2-20x-24\le 0, divide that annoying 4 out:

x^2-5x-6\le 0

factor:

(x-6)(x+1)\le 0

From here it's easy.

It's a matter of opinion, but I think my way is easier.


Cases are almost never easier, this way it is absolutely immediately clear what the intervals are. And in any case it's senseless to talk about it, both are available to the op to choose as he likes, this isn't a popularity of solutions contest.

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