Solve the equation- 2x2 - 3x - 9 - 0

The correct option is B 3 ≤ x ≤32 We have, 2x^2 + 3x 9 ≤ 0 ⟹ 2x^2 + 6x 3x 9 ≤ 0 ⟹ (x + 3) (2x 3) ≤ 0 ⟹ #160; 3 ≤ x ≤32 ...

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Equation

Solve the equ...

Question

Solve the equation: $2 x^{2} + 3 x - 9 \leq 0$

- \frac{3}{2} \leq x \leq 3

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- 3 \leq x \leq \frac{3}{2}

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x \in (- \infty, - 3) \cup (\frac{3}{2}, \infty)

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x \in (- \infty, - \frac{3}{2}) \cup (3, \infty)

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Solve the following equation:

$\frac{2 x}{x - 3} + \frac{1}{2 x + 3} + \frac{3 x + 9}{(x - 3) (2 x + 3)} = 0; x \neq 3, x \neq - \frac{3}{2}$

(x + y)^{2 / 3} + 2 (x - y)^{2 / 3} = 3 (x^{2} - y^{2})^{1 / 3}, 3 x - 2 y = 13

\frac{2 x^{3} - 3 x^{2} + x + 1}{2 x^{3} - 3 x^{2} - x - 1} = \frac{3 x^{3} - x^{2} + 5 x - 13}{3 x^{3} - x^{2} - 5 x + 13}

\sqrt{3} x^{2} - 2 \sqrt{3} x + 3 \sqrt{3} = 0

\frac{3^{x} + 3 - x}{3^{x} - 3 - x} = 2.

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