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# Help. Thanks. :D

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Solve the inequality \$x^3  + 4x > 5x^2\$.

May 11, 2020

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To solve the inequality:  x3 + 4x  >  5x2

x3 - 5x2 + 4x  > 0

x(x2 - 5x + 4)  >  0

x(x -1)(x - 4)  >  0

The points that divide the number line into regions are  0,  1,  and  4.

[None of these points are possible answers because the problem has  >  not  >=.]

We need to check one point from each of these regions:  x < 0,    0 < x < 1,   1 < x < 4,   x > 4

Region:  x < 0:  We can use any point, I will choose -5 and place this number into  x(x -1)(x - 4)  >  0.

-5  makes x negative;  -5 makes (x - 1) negative;  -5 makes (x - 4) negative;  multiplying three negatives together,

we get a negative number, not the positive that the problem says. This region doesn't work.

Region:  0 < x < 1:  We can use any point; I will choose ½.

½  makes x positive;   ½  makes (x - 1) negative;   ½  makes (x - 4) negative; multiplying a positive times a negative

times another negative gives us a positive number, just what we need. This region works.

Region:  1 < x < 4:  We can use any point, I will choose 2.

2  makes x positive;  2 makes (x - 1) positive;  2 makes (x - 4) negative; multiplying a positive times a positive

times a negative gives us a negative number. This region doesn't work.

Region:  x > 4:  We an use any point, I will choose 10.

10  makes x positive;  10 makes (x - 1) positive;  10 makes (x - 4) positive; multiply all positive numbers together

gives us a positive number. This region works.

The answer consists of the regions that work.  0 < x < 1   and   x > 4.

May 11, 2020