Solve 2x² + 3x - 2 > 0 - Edexcel - GCSE Maths - Question 19 - 2017 - Paper 3

Setting each factor to zero, we find the roots:

1. $2x - 1 = 0 \implies x = \frac{1}{2}$
2. $x + 2 = 0 \implies x = -2$

The x-values that make the expression zero are $x = -2$ and $x = \frac{1}{2}$. These roots divide the number line into three intervals:

1. $(-\infty, -2)$
2. $(-2, \frac{1}{2})$
3. $(\frac{1}{2}, \infty)$

We test a value from each interval in the inequality $2x^2 + 3x - 2 > 0$:

1. For $x = -3$ (from $(-\infty, -2)$): $2(-3)^2 + 3(-3) - 2 = 18 - 9 - 2 = 7 > 0$

2. For $x = 0$ (from $(-2, \frac{1}{2})$): $2(0)^2 + 3(0) - 2 = -2 < 0$

3. For $x = 1$ (from $(\frac{1}{2}, \infty)$): $2(1)^2 + 3(1) - 2 = 2 + 3 - 2 = 3 > 0$

The expression is greater than zero in the intervals $(-\infty, -2)$ and $(\frac{1}{2}, \infty)$. Therefore, the solution is:

$x < -2 \quad \text{or} \quad x > \frac{1}{2}$