Question

solve $x^2 - 8x + 15 < 0$.
recall that the quadratic factors as:
$(x - 3)(x - 5) < 0$
therefore, the intervals that must be tested are $x < 3, 3 < x < 5$ and $x > 5$.
the solution set for the quadratic inequality is:
$\bigcirc$ $(-\infty, 3)$
$\bigcirc$ $(-\infty, 3)$ or $(5, \infty)$
$\bigcirc$ $(3, 5)$
$\bigcirc$ $(5, \infty)$

Explanation:

Step1: Analyze the factored form

We have the inequality \((x - 3)(x - 5)<0\). A product of two numbers is negative when one number is positive and the other is negative.

Step2: Test the intervals

• For \(x < 3\), let's take \(x = 0\). Then \((0 - 3)(0 - 5)=(- 3)\times(-5) = 15>0\), so this interval does not satisfy the inequality.
• For \(3 < x < 5\), let's take \(x = 4\). Then \((4 - 3)(4 - 5)=(1)\times(-1)=- 1<0\), so this interval satisfies the inequality.
• For \(x>5\), let's take \(x = 6\). Then \((6 - 3)(6 - 5)=(3)\times(1) = 3>0\), so this interval does not satisfy the inequality.

Answer:

\((3,5)\)