Question

Which values of x are satisfied by the inequality 2x² + x-3 <0?

• A. $-(\frac{3}{2})<x<1$
• B. $-1<x<(\frac{2}{3})$
• C. x $>$ 1
• D. $x< -(\frac{2}{5})$

Answer 1

The Correct Option is A

The correct option is (A): $-(\frac{3}{2})<x<1$
Explanation: The inequality given is:
$$2x^2 + x - 3 < 0$$
To solve this quadratic inequality, we first solve the corresponding equation:
$$2x^2 + x - 3 = 0$$
We can use the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Here, $a = 2$, $b = 1$, and $c = -3$. Substituting into the quadratic formula:
$$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-3)}}{2(2)}$$
$$x = \frac{-1 \pm \sqrt{1 + 24}}{4}$$
$$x = \frac{-1 \pm \sqrt{25}}{4}$$
$$x = \frac{-1 \pm 5}{4}$$
So, the solutions are:

$$x = \frac{-1 + 5}{4} = 1 \quad \text{and} \quad x = \frac{-1 - 5}{4} = -\frac{3}{2}$$
These are the critical points. Now, we check the sign of $2x^2 + x - 3$ in the intervals divided by these points: $(-∞, -\frac{3}{2})$, $(- \frac{3}{2}, 1)$, and $(1, ∞)$.
1. For $x < -\frac{3}{2}$, the expression is positive.
2. For $-\frac{3}{2} < x < 1$, the expression is negative.
3. For $x > 1$, the expression is positive.
Thus, the inequality $2x^2 + x - 3 < 0$ is satisfied when:
$$-\frac{3}{2} < x < 1$$

So, the correct answer is A: $-\frac{3}{2} < x < 1$.