Solve the inequality (2 - x)(2x - 3) > 0.

Step-by-step solution

1. 1

   Leave it factored and find the zeros. A product's sign is decided by its factors, so do not expand:

   $2-x=0\Rightarrow x=2,\qquad 2x-3=0\Rightarrow x=\frac32$

   These split the line into $\left(-\infty,\tfrac32\right)$, $\left(\tfrac32,2\right)$ and $(2,\infty)$.

2. 2

   Test one point in each interval. At $x=0$: $(2)(-3)=-6<0$. At $x=1.75$ (between the roots): $(0.25)(0.5)=0.125>0$. At $x=3$: $(-1)(3)=-3<0$. The pattern is negative, positive, negative.

3. 3

   Explain the pattern from the leading coefficient. If you did expand, the leading term would be $(-x)(2x)=-2x^2$ — a downward parabola. Downward parabolas are positive between their roots and negative outside, the mirror image of the familiar case. Writing "outside the roots" out of habit is the standard error here.

4. 4

   Take the interval between the roots, noting which root is smaller. Since $\tfrac32=1.5<2$:

   $\frac32<x<2$

5. 5

   Exclude the endpoints. The inequality is strict, so the two points where the product is exactly $0$ are not in the solution set. In interval notation, $\left(\tfrac32,2\right)$ — an open interval of width just $\tfrac12$.

6. 6

   Verify by scanning. Evaluating the product with exact fractions at $10{,}001$ points on $[-5,5]$ and comparing against $\left(\tfrac32,2\right)$ gives agreement at every point ✓.