Answer :
To solve the inequality \( f(x) = x^2 + 2x - 15 \leq 0 \), we will analyze the quadratic function \( f(x) \) and determine where it is less than or equal to zero. Here's the step-by-step process:
1. Identify the quadratic function and the inequality:
[tex]\[ f(x) = x^2 + 2x - 15 \][/tex]
We need to find the values of \( x \) for which \( f(x) \leq 0 \).
2. Find the roots of the quadratic equation:
The roots of the equation \( x^2 + 2x - 15 = 0 \) can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -15 \).
Calculate the discriminant:
[tex]\[ b^2 - 4ac = 2^2 - 4(1)(-15) = 4 + 60 = 64 \][/tex]
Now, find the roots:
[tex]\[ x = \frac{-2 \pm \sqrt{64}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-2 \pm 8}{2} \][/tex]
This results in two roots:
[tex]\[ x = \frac{-2 + 8}{2} = \frac{6}{2} = 3 \][/tex]
[tex]\[ x = \frac{-2 - 8}{2} = \frac{-10}{2} = -5 \][/tex]
Therefore, the roots of the quadratic equation are \( x = -5 \) and \( x = 3 \).
3. Determine the intervals to test:
The roots \( x = -5 \) and \( x = 3 \) divide the real number line into three intervals:
- \( (-\infty, -5) \)
- \( (-5, 3) \)
- \( (3, \infty) \)
4. Test the sign of the function in each interval:
- For \( x \in (-\infty, -5) \), choose a test point such as \( x = -6 \):
[tex]\[ f(-6) = (-6)^2 + 2(-6) - 15 = 36 - 12 - 15 = 9 \][/tex]
Since \( 9 > 0 \), \( f(x) > 0 \) in \( (-\infty, -5) \).
- For \( x \in (-5, 3) \), choose a test point such as \( x = 0 \):
[tex]\[ f(0) = (0)^2 + 2(0) - 15 = -15 \][/tex]
Since \( -15 \leq 0 \), \( f(x) \leq 0 \) in \( (-5, 3) \).
- For \( x \in (3, \infty) \), choose a test point such as \( x = 4 \):
[tex]\[ f(4) = (4)^2 + 2(4) - 15 = 16 + 8 - 15 = 9 \][/tex]
Since \( 9 > 0 \), \( f(x) > 0 \) in \( (3, \infty) \).
5. Conclusion:
Based on the test points, the quadratic function \( f(x) \) is less than or equal to zero in the interval \( [-5, 3] \).
Therefore, the solution to the inequality \( x^2 + 2x - 15 \leq 0 \) is:
[tex]\[ x \in [-5, 3] \][/tex]
This means that the inequality holds for all [tex]\( x \)[/tex] in the interval from [tex]\(-5\)[/tex] to [tex]\(3\)[/tex], inclusive.
1. Identify the quadratic function and the inequality:
[tex]\[ f(x) = x^2 + 2x - 15 \][/tex]
We need to find the values of \( x \) for which \( f(x) \leq 0 \).
2. Find the roots of the quadratic equation:
The roots of the equation \( x^2 + 2x - 15 = 0 \) can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -15 \).
Calculate the discriminant:
[tex]\[ b^2 - 4ac = 2^2 - 4(1)(-15) = 4 + 60 = 64 \][/tex]
Now, find the roots:
[tex]\[ x = \frac{-2 \pm \sqrt{64}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-2 \pm 8}{2} \][/tex]
This results in two roots:
[tex]\[ x = \frac{-2 + 8}{2} = \frac{6}{2} = 3 \][/tex]
[tex]\[ x = \frac{-2 - 8}{2} = \frac{-10}{2} = -5 \][/tex]
Therefore, the roots of the quadratic equation are \( x = -5 \) and \( x = 3 \).
3. Determine the intervals to test:
The roots \( x = -5 \) and \( x = 3 \) divide the real number line into three intervals:
- \( (-\infty, -5) \)
- \( (-5, 3) \)
- \( (3, \infty) \)
4. Test the sign of the function in each interval:
- For \( x \in (-\infty, -5) \), choose a test point such as \( x = -6 \):
[tex]\[ f(-6) = (-6)^2 + 2(-6) - 15 = 36 - 12 - 15 = 9 \][/tex]
Since \( 9 > 0 \), \( f(x) > 0 \) in \( (-\infty, -5) \).
- For \( x \in (-5, 3) \), choose a test point such as \( x = 0 \):
[tex]\[ f(0) = (0)^2 + 2(0) - 15 = -15 \][/tex]
Since \( -15 \leq 0 \), \( f(x) \leq 0 \) in \( (-5, 3) \).
- For \( x \in (3, \infty) \), choose a test point such as \( x = 4 \):
[tex]\[ f(4) = (4)^2 + 2(4) - 15 = 16 + 8 - 15 = 9 \][/tex]
Since \( 9 > 0 \), \( f(x) > 0 \) in \( (3, \infty) \).
5. Conclusion:
Based on the test points, the quadratic function \( f(x) \) is less than or equal to zero in the interval \( [-5, 3] \).
Therefore, the solution to the inequality \( x^2 + 2x - 15 \leq 0 \) is:
[tex]\[ x \in [-5, 3] \][/tex]
This means that the inequality holds for all [tex]\( x \)[/tex] in the interval from [tex]\(-5\)[/tex] to [tex]\(3\)[/tex], inclusive.
