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Method of Intervals

The solution ...

Question

The solution set of $x^{3} - 10 x^{2} + 21 x > 0$ is

(0, 3)

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(0, \infty)

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(3, 7)

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(0, 3) \cup (7, \infty)

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Solution

The correct option is D $(0, 3) \cup (7, \infty)$
$x^{3} - 10 x^{2} + 21 x > 0 \Rightarrow x (x - 3) (x - 7) > 0$

$∴ x \in (0, 3) \cup (7, \infty)$

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Similar questions

x^{3} - 10 x^{2} + 21 x > 0

r

^{69} C_{3 r - 1} -^{69} C_{r^{2}} =^{69} C_{r^{2} - 1} -^{69} C_{3 r}

x_{1 / 10} (x^{2} + x + 1) > 0

x - 2 > 0, x - 9 < 0

x^{2} - 9 \geq 0

x^{3} + 21 x + 342 = 0

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