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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given inequality: $2x^2 + 3x - 9 \le 0$Factorizing the quadratic expression: $2x^2 + 6x - 3x - 9 \le 0$$2x(x + 3) - 3(x + 3) \le 0$$(2x - 3)(x + 3) \le 0$The roots of the equation $2x^2 + 3x - 9 = 0$ are $x = 3/2$ and $x = -3$.For a quadratic expression $ax^2 + bx + c \le 0$ where $a > 0$,the solution lies between the roots,i.e.,$	ext{root}_1 \le x \le 	ext{root}_2$.Thus,the solution is $-3 \le x \le 3/2$.

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

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