Solving Linear Inequalities: x^2x - 4x + 8 < x(x + 8)

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DavidBr
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Solving Linear Inequalities: x^2x - 4x + 8 < x(x + 8)

Postby DavidBr » Sun May 03, 2009 10:10 pm

Ok, so I'm working on a problem and I'm seeing two ways of doing the problem. Trouble is, neither way comes up with the same result. Advice?

First Route:
x^2 -4x +8 < x(x+8)
x^2 -4x +8 < x^2 +8x

Now the trouble starts:

x^2 -x^2 -4x +4x +8 < x^2 -x^2 +8x +4x
8 < 12x
8/12 < 12/12 x
(2/3) < x

...0, ((2/3), 1, 2, 3, ...+inf.

((2/3),+inf.)


Second route:
x^2 -4x +8 < x(x+8)
x^2 -4x +8 < x^2 +8x

Alternate:

x^2 -x^2 -4x -8x +8 -8 < x^2 -x^2 +8x -8x -8
4x < -8
x < -2


-inf.... -4, -3, -2,) -1, 0, ...

(-inf.,-2)


Now I know having 4x be less than -8 would in most cases result in an incorrect assessment of the equation. However, this situation begs the question, "Is there a convention for determining what gets moved in what order?".

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stapel_eliz
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Re: Solving Linear Inequalities: x^2x - 4x + 8 < x(x + 8)

Postby stapel_eliz » Sun May 03, 2009 11:36 pm

It should be noted that these are not, strictly-speaking, "linear" inequalities, but quadratic inequalities. :oops:

DavidBr wrote:First Route:
x^2 -4x +8 < x(x+8)
x^2 -4x +8 < x^2 +8x

Get everything together on one side:

. . . . .

Then, since this has simplified to a linear inequality, proceed as usual:

. . . . .

. . . . .

So the solution is the interval:

. . . . .

DavidBr wrote:Alternate:
x^2 -x^2 -4x -8x +8 -8 < x^2 -x^2 +8x -8x -8
4x < -8

But -4x - 8x does not equal 4x; it equals -12x. :wink:

DavidBr
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Re: Solving Linear Inequalities: x^2x - 4x + 8 < x(x + 8)

Postby DavidBr » Mon May 04, 2009 1:30 am

Thank you for the reply.

To be honest, I don't know the difference yet between Linear an Quadratic Inequalities. I'm in the first month of a math 90 course and stumbling along. And now that you've pointed out that basic math error, I'm also in Kicking my-self 101! I should have know it was an error that was causing the problems. Thank you for the heads up and the help.


-David


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