To solve the inequality [tex]\(x^2 - 8x + 15 < 0\)[/tex] and find the critical points, follow these steps:
1. Identify the quadratic equation:
The inequality we need to solve is:
[tex]\[
x^2 - 8x + 15 < 0
\][/tex]
To proceed, we first solve the related quadratic equation:
[tex]\[
x^2 - 8x + 15 = 0
\][/tex]
2. Find the roots of the quadratic equation:
We need to factor the quadratic equation or use the quadratic formula to find its roots. The quadratic formula for an equation [tex]\(ax^2 + bx + c = 0\)[/tex] is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For the equation [tex]\(x^2 - 8x + 15 = 0\)[/tex], we identify [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 15 \)[/tex].
Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = (-8)^2 - 4 \cdot 1 \cdot 15 = 64 - 60 = 4
\][/tex]
Now compute the roots:
[tex]\[
x = \frac{-(-8) \pm \sqrt{4}}{2 \cdot 1} = \frac{8 \pm 2}{2}
\][/tex]
This gives us the roots:
[tex]\[
x_1 = \frac{8 + 2}{2} = \frac{10}{2} = 5
\][/tex]
[tex]\[
x_2 = \frac{8 - 2}{2} = \frac{6}{2} = 3
\][/tex]
3. Determine the intervals for the inequality:
The roots divide the number line into three intervals:
[tex]\[
(-\infty, 3), \quad (3, 5), \quad (5, \infty)
\][/tex]
To determine where the inequality [tex]\(x^2 - 8x + 15 < 0\)[/tex] holds, test points within each interval:
- For an interval like [tex]\((- \infty, 3)\)[/tex], we can pick [tex]\(x = 0\)[/tex]. Substituting [tex]\(x = 0\)[/tex] into the quadratic expression gives:
[tex]\[
0^2 - 8 \cdot 0 + 15 = 15 \quad (\text{which is greater than 0})
\][/tex]
- For the interval [tex]\((3, 5)\)[/tex], we can pick [tex]\(x = 4\)[/tex]. Substituting [tex]\(x = 4\)[/tex] gives:
[tex]\[
4^2 - 8 \cdot 4 + 15 = 16 - 32 + 15 = -1 \quad (\text{which is less than 0})
\][/tex]
- For the interval [tex]\((5, \infty)\)[/tex], we can pick [tex]\(x = 6\)[/tex]. Substituting [tex]\(x = 6\)[/tex] gives:
[tex]\[
6^2 - 8 \cdot 6 + 15 = 36 - 48 + 15 = 3 \quad (\text{which is greater than 0})
\][/tex]
This shows that the quadratic expression is negative only in the interval [tex]\( (3, 5) \)[/tex].
4. Conclusion:
The inequality [tex]\(x^2 - 8x + 15 < 0\)[/tex] holds for [tex]\( x \)[/tex] in the interval [tex]\((3, 5)\)[/tex].
Thus, the critical points for the inequality are:
[tex]\[ \boxed{3 \text{ and } 5} \][/tex]
