Solve- 2 x 2 -3x-9- 0

The correct option is C 3≤ x≤32 2x^2 + 3x 9 ≤ 0 ⇒ (2) (3) (x + 3) ≤ 0 At x lt; 3 ; (2x 3)(x + 3) gt; 0 ⇒ 3 ≤ x ≤32 ( ...

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Byju's Answer

Standard XII

Mathematics

Method of Intervals

Solve: 2 x ...

Question

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

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

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

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

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None of these

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Solution

The correct option is C $- 3 \leq x \leq \frac{3}{2}$
$2 x^{2} + 3 x - 9 \leq 0$
$\Rightarrow (2) (3) (x + 3) \leq 0$
At $x < - 3; (2 x - 3) (x + 3) > 0$
$\Rightarrow - 3 \leq x \leq \frac{3}{2}$
( $∵ (2 x - 3) & (x + 3)$ has odd power $∴$ sign changes at $- 3 & 3 / 2$ )

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$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$