Connect with knowledgeable experts and enthusiasts on IDNLearn.com. Our platform provides detailed and accurate responses from experts, helping you navigate any topic with confidence.
Sagot :
To solve the inequality [tex]\( f(x) = x^2 + 4x - 45 \geq 0 \)[/tex], we need to find the regions where the function [tex]\( f(x) \)[/tex] is greater than or equal to zero.
### Step-by-Step Solution:
1. Find the roots of the quadratic equation:
To determine where the function changes sign, first find the roots of the quadratic equation [tex]\( f(x) = x^2 + 4x - 45 = 0 \)[/tex]. These are the points where the function intersects the x-axis.
2. Solve the quadratic equation [tex]\( x^2 + 4x - 45 = 0 \)[/tex]:
This can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are:
[tex]\[ a = 1, \quad b = 4, \quad c = -45 \][/tex]
Plugging in the values, we get:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1} \][/tex]
This simplifies to:
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 180}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{196}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 14}{2} \][/tex]
Thus, we have two roots:
[tex]\[ x = \frac{-4 + 14}{2} = 5 \][/tex]
[tex]\[ x = \frac{-4 - 14}{2} = -9 \][/tex]
3. Critical points and test intervals:
The roots [tex]\( x = -9 \)[/tex] and [tex]\( x = 5 \)[/tex] are critical points. We will test the intervals around these points to determine where the function is non-negative.
The intervals to test are:
- [tex]\( x \leq -9 \)[/tex]
- [tex]\( -9 \leq x \leq 5 \)[/tex]
- [tex]\( x \geq 5 \)[/tex]
4. Test the intervals:
- For [tex]\( x \leq -9 \)[/tex]:
Choose a point in this interval, such as [tex]\( x = -10 \)[/tex]:
[tex]\[ f(-10) = (-10)^2 + 4(-10) - 45 = 100 - 40 - 45 = 15 \][/tex]
Since [tex]\( f(-10) > 0 \)[/tex], [tex]\( f(x) \)[/tex] is non-negative in the interval [tex]\( x \leq -9 \)[/tex].
- For [tex]\( -9 \leq x \leq 5 \)[/tex]:
Choose a point in this interval, such as [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 + 4(0) - 45 = -45 \][/tex]
Since [tex]\( f(0) < 0 \)[/tex], [tex]\( f(x) \)[/tex] is negative in the interval [tex]\( -9 \leq x \leq 5 \)[/tex].
- For [tex]\( x \geq 5 \)[/tex]:
Choose a point in this interval, such as [tex]\( x = 6 \)[/tex]:
[tex]\[ f(6) = 6^2 + 4(6) - 45 = 36 + 24 - 45 = 15 \][/tex]
Since [tex]\( f(6) > 0 \)[/tex], [tex]\( f(x) \)[/tex] is non-negative in the interval [tex]\( x \geq 5 \)[/tex].
5. Conclusion:
The function [tex]\( f(x) = x^2 + 4x - 45 \)[/tex] is non-negative in the intervals where [tex]\( x \leq -9 \)[/tex] and [tex]\( x \geq 5 \)[/tex]. Thus, the solution to the inequality [tex]\( f(x) \geq 0 \)[/tex] is:
[tex]\[ x \in (-\infty, -9] \cup [5, \infty) \][/tex]
Therefore, the regions where the quadratic function is greater than or equal to zero are [tex]\( x \leq -9 \)[/tex] and [tex]\( x \geq 5 \)[/tex].
### Step-by-Step Solution:
1. Find the roots of the quadratic equation:
To determine where the function changes sign, first find the roots of the quadratic equation [tex]\( f(x) = x^2 + 4x - 45 = 0 \)[/tex]. These are the points where the function intersects the x-axis.
2. Solve the quadratic equation [tex]\( x^2 + 4x - 45 = 0 \)[/tex]:
This can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are:
[tex]\[ a = 1, \quad b = 4, \quad c = -45 \][/tex]
Plugging in the values, we get:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1} \][/tex]
This simplifies to:
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 180}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{196}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 14}{2} \][/tex]
Thus, we have two roots:
[tex]\[ x = \frac{-4 + 14}{2} = 5 \][/tex]
[tex]\[ x = \frac{-4 - 14}{2} = -9 \][/tex]
3. Critical points and test intervals:
The roots [tex]\( x = -9 \)[/tex] and [tex]\( x = 5 \)[/tex] are critical points. We will test the intervals around these points to determine where the function is non-negative.
The intervals to test are:
- [tex]\( x \leq -9 \)[/tex]
- [tex]\( -9 \leq x \leq 5 \)[/tex]
- [tex]\( x \geq 5 \)[/tex]
4. Test the intervals:
- For [tex]\( x \leq -9 \)[/tex]:
Choose a point in this interval, such as [tex]\( x = -10 \)[/tex]:
[tex]\[ f(-10) = (-10)^2 + 4(-10) - 45 = 100 - 40 - 45 = 15 \][/tex]
Since [tex]\( f(-10) > 0 \)[/tex], [tex]\( f(x) \)[/tex] is non-negative in the interval [tex]\( x \leq -9 \)[/tex].
- For [tex]\( -9 \leq x \leq 5 \)[/tex]:
Choose a point in this interval, such as [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 + 4(0) - 45 = -45 \][/tex]
Since [tex]\( f(0) < 0 \)[/tex], [tex]\( f(x) \)[/tex] is negative in the interval [tex]\( -9 \leq x \leq 5 \)[/tex].
- For [tex]\( x \geq 5 \)[/tex]:
Choose a point in this interval, such as [tex]\( x = 6 \)[/tex]:
[tex]\[ f(6) = 6^2 + 4(6) - 45 = 36 + 24 - 45 = 15 \][/tex]
Since [tex]\( f(6) > 0 \)[/tex], [tex]\( f(x) \)[/tex] is non-negative in the interval [tex]\( x \geq 5 \)[/tex].
5. Conclusion:
The function [tex]\( f(x) = x^2 + 4x - 45 \)[/tex] is non-negative in the intervals where [tex]\( x \leq -9 \)[/tex] and [tex]\( x \geq 5 \)[/tex]. Thus, the solution to the inequality [tex]\( f(x) \geq 0 \)[/tex] is:
[tex]\[ x \in (-\infty, -9] \cup [5, \infty) \][/tex]
Therefore, the regions where the quadratic function is greater than or equal to zero are [tex]\( x \leq -9 \)[/tex] and [tex]\( x \geq 5 \)[/tex].
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.