## 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)

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?".

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

It should be noted that these are not, strictly-speaking, "linear" inequalities, but quadratic inequalities.
First Route:
x^2 -4x +8 < x(x+8)
x^2 -4x +8 < x^2 +8x
Get everything together on one side:

. . . . .$0\, <\, 12x\, -\, 8$

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

. . . . .$8\, <\, 12x$

. . . . .$\frac{2}{3}\, <\, x$

So the solution is the interval:

. . . . .$\left(\frac{2}{3},\, \infty)$
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.

DavidBr
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