Solve the inequality 2x2 - 3x - 2 > 0.

Step-by-step solution

1. 1

   Confirm the inequality is already in standard form. Everything is on the left with $0$ on the right, so we can go straight to factoring. The leading coefficient is $2$, which means the factorisation is non-monic and the split must account for it.

2. 2

   Factor the non-monic quadratic. Look for integers whose product is $ac = 2\cdot(-2) = -4$ and whose sum is $b=-3$: those are $-4$ and $1$. Splitting the middle term, $2x^2-4x+x-2 = 2x(x-2)+1(x-2)$, so $2x^2-3x-2 = (2x+1)(x-2).$ Expanding back gives $2x^2-4x+x-2=2x^2-3x-2$, which checks.

3. 3

   Find the critical points. Set each factor to zero: $2x+1=0$ gives $x=-\tfrac12$, and $x-2=0$ gives $x=2$. These are the only places the expression can change sign.

4. 4

   Test the sign on each interval. The points split the line into $\left(-\infty,-\tfrac12\right)$, $\left(-\tfrac12,2\right)$ and $(2,\infty)$. At $x=-1$: $(2(-1)+1)((-1)-2) = (-1)(-3) = 3>0$. At $x=0$: $(1)(-2) = -2<0$. At $x=3$: $(7)(1) = 7>0$. The pattern $+,-,+$ is exactly what an upward-opening parabola must do.

5. 5

   Keep the intervals where the product is positive. Because the inequality is strict, the roots themselves are excluded: $x<-\tfrac12 \quad \text{or} \quad x>2, \qquad \text{i.e. } \left(-\infty,-\tfrac12\right)\cup(2,\infty).$

6. 6

   Remember the shortcut. For $a>0$ and two distinct real roots $r_1<r_2$, the solution of $ax^2+bx+c>0$ is always outside the roots and of $<0$ always between them - the sign chart above just confirms it.