Solve the polynomial inequality. Express the solution set in in interval notation. $13 x^{2} - 5 x \leq 0$

[- \infty, 0] \cup [\frac{5}{13}, \infty]

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[0, \frac{5}{13}]

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[0, \frac{13}{5}]

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[- \frac{5}{13}, 0]

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Solution

The correct option is B $[0, \frac{5}{13}]$
Given, $13 x^{2} - 5 x \leq 0$
$\Rightarrow x (13 x - 5) \leq 0$
$\Rightarrow x \in [0, \frac{5}{13}]$
Option B is correct.

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