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: \( -\frac{3}{2} < x < 1 \).