Solve the equation: $2x^2 + 3x - 9 \leq 0$
- $ -\dfrac{3}{2} \leq x \leq 3$
- $ -3 \leq x \leq \dfrac{3}{2}$
- $ x \in \left(-\infty,-3\right) \cup \left(\dfrac{3}{2},\infty\right)$
- $ x \in \left(-\infty,-\dfrac{3}{2}\right) \cup (3,\infty)$
Reveal answer
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B
Correct answer
Explanation
2x^2 + 3x - 9 <= 0. Factors: (2x - 3)(x + 3) <= 0. Roots are -3 and 1.5. The inequality holds between the roots: -3 <= x <= 1.5.
AI explanation
To solve the inequality 2x squared plus 3x minus 9 is less than or equal to 0, we first find the roots of the corresponding equation by factoring it into 2x minus 3 times x plus 3 equals 0. This gives the critical points x equals 3 halves and x equals negative 3. Since the coefficient of x squared is positive, the parabola opens upwards, meaning the expression is less than or equal to zero between the roots, so x is greater than or equal to negative 3 and less than or equal to 3 halves.