Question

solve the following inequality
$x^2 - 13x + 30 < 0$

select the correct choice below and, if necessary, fill in the answer box.

a. the solution set is \boxed{}.
(type your answer in interval notation. use integers or fractions for any numbers in the expression.)

b. there is no real solution.

Explanation:

Step1: Factor the quadratic

We factor \(x^{2}-13x + 30\). We need two numbers that multiply to \(30\) and add to \(- 13\). The numbers are \(-3\) and \(-10\). So, \(x^{2}-13x + 30=(x - 3)(x - 10)\).

Step2: Find critical points

Set each factor equal to zero: \(x-3 = 0\) gives \(x = 3\); \(x - 10=0\) gives \(x = 10\). These are the critical points that divide the number line into intervals: \((-\infty,3)\), \((3,10)\), and \((10,\infty)\).

Step3: Test intervals

• For \(x<3\) (e.g., \(x = 0\)): \((0 - 3)(0 - 10)=(-3)\times(-10)=30>0\), so the inequality \(x^{2}-13x + 30<0\) is false here.
• For \(3
• For \(x>10\) (e.g., \(x = 11\)): \((11 - 3)(11 - 10)=(8)\times(1)=8>0\), so the inequality is false here.

Answer:

A. The solution set is \((3,10)\)