Problem
Find solution set (over the real numbers) of the quadratic inequality \(x^2 - 5x + 6 \le 0\).
Step 1: Factor the quadratic
Factor by finding two numbers whose product is \(6\) and whose sum is \(-5\): \(-2\) and \(-3\). Therefore,
\[ x^2 - 5x + 6 = (x - 2)\cdot(x - 3). \]
Step 2: Identify critical points
The factors are zero at \(x = 2\) and \(x = 3\). These points split the real line into intervals: \((-\infty,2)\), \((2,3)\), and \((3,\infty)\).
Step 3: Sign analysis on each interval
The product \((x-2)\cdot(x-3)\) changes sign only at the critical points. A sign chart determines where the product is \(\le 0\).
| Interval | Test value | \(x-2\) | \(x-3\) | \((x-2)\cdot(x-3)\) | Satisfies \(\le 0\)? |
|---|---|---|---|---|---|
| \((-\infty,2)\) | \(x=0\) | \(0-2<0\) | \(0-3<0\) | \((-) \cdot (-) = (+)\) | No |
| \((2,3)\) | \(x=2.5\) | \(2.5-2>0\) | \(2.5-3<0\) | \((+) \cdot (-) = (-)\) | Yes |
| \((3,\infty)\) | \(x=4\) | \(4-2>0\) | \(4-3>0\) | \((+) \cdot (+) = (+)\) | No |
Because the inequality is \(\le 0\), the endpoints where the expression equals \(0\) are included: \(x=2\) and \(x=3\).
Solution set
Interval notation: \([2,3]\)
Set-builder notation: \(\{x \in \mathbb{R} \mid 2 \le x \le 3\}\)
Visualization: solution set on a number line
Common checks and pitfalls
- The symbol \(\le\) means endpoints are included; using open circles would be incorrect for this problem.
- A quick verification: \(x=2.5\) gives \((2.5-2)\cdot(2.5-3) < 0\), so values between \(2\) and \(3\) belong to the solution set.
- Values outside \([2,3]\) make both factors the same sign, producing a positive product, which fails the \(\le 0\) condition.