The solution of the equation 2x^2 + 3x - 9 le | vedclass

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Question

(B) Given the quadratic inequality: $2x^2 + 3x - 9 \le 0$Step $1$: Factorize the quadratic expression.$2x^2 + 6x - 3x - 9 \le 0$$2x(x + 3) - 3(x + 3)

Vedclass · Accepted Answer

$-3 \le x \le 3/2$

Vedclass · Answer

(B) Given the quadratic inequality: $2x^2 + 3x - 9 \le 0$Step $1$: Factorize the quadratic expression.$2x^2 + 6x - 3x - 9 \le 0$$2x(x + 3) - 3(x + 3) \le 0$$(2x - 3)(x + 3) \le 0$Step $2$: Find the critical points by setting the factors to zero.$2x - 3 = 0 \implies x = 3/2$$x + 3 = 0 \implies x = -3$Step $3$: Determine the interval where the product is less than or equal to zero.For a quadratic $(x - \alpha)(x - \beta) \le 0$ where $\alpha Here,$-3 \le x \le 3/2$.Thus,the correct option is $B$.

The solution of the equation $2x^2 + 3x - 9 \le 0$ is given by

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