- #1

- 32

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1296(b^3) - 324(b^2) - 1008b + 108 > 0.

I want to know for what values of b this inequality is true. Any suggestions?

- Thread starter lokisapocalypse
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- #1

- 32

- 0

1296(b^3) - 324(b^2) - 1008b + 108 > 0.

I want to know for what values of b this inequality is true. Any suggestions?

- #2

ehild

Homework Helper

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I do not think you can simplify the expression too mutch except factoring out 36.lokisapocalypse said:

1296(b^3) - 324(b^2) - 1008b + 108 > 0.

I want to know for what values of b this inequality is true. Any suggestions?

Anyway, it is a third order polynomial, which goes from -infinity to + infinity, and having either one or three real roots. You can find them approximately(-0.821, 0.105, 0.966). As the polynomial is negative for very low values of b and positive for very high ones, it is positive between the first two roots and also for b higher than the highest root.

ehild

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