Join IDNLearn.com and start getting the answers you've been searching for. Join our Q&A platform to receive prompt and accurate responses from knowledgeable professionals in various fields.
Sagot :
Let's solve the polynomial inequality [tex]\( x^2 - 12x + 27 > 0 \)[/tex] step by step.
### Step 1: Set the polynomial equal to zero
Firstly, we need to find the points where the polynomial [tex]\( x^2 - 12x + 27 \)[/tex] equals zero. These points are called boundary points and can be found by solving the equation:
[tex]\[ x^2 - 12x + 27 = 0 \][/tex]
### Step 2: Solve the quadratic equation
To solve the quadratic equation, we can factorize it:
[tex]\[ x^2 - 12x + 27 = (x - 3)(x - 9) = 0 \][/tex]
Setting each factor to zero gives us the boundary points:
[tex]\[ x - 3 = 0 \implies x = 3 \][/tex]
[tex]\[ x - 9 = 0 \implies x = 9 \][/tex]
So, the boundary points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 9 \)[/tex].
### Step 3: Determine the sign intervals
The boundary points divide the number line into three intervals:
1. [tex]\( (-\infty, 3) \)[/tex]
2. [tex]\( (3, 9) \)[/tex]
3. [tex]\( (9, \infty) \)[/tex]
We need to determine whether [tex]\( x^2 - 12x + 27 \)[/tex] is positive or negative in each of these intervals.
#### Interval [tex]\( (-\infty, 3) \)[/tex]:
Choose a test point less than 3, for example, [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 12(0) + 27 = 27 \][/tex]
Since 27 is positive, [tex]\( f(x) > 0 \)[/tex] on the interval [tex]\( (-\infty, 3) \)[/tex].
#### Interval [tex]\( (3, 9) \)[/tex]:
Choose a test point between 3 and 9, for example, [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 5^2 - 12(5) + 27 = 25 - 60 + 27 = -8 \][/tex]
Since -8 is negative, [tex]\( f(x) < 0 \)[/tex] on the interval [tex]\( (3, 9) \)[/tex].
#### Interval [tex]\( (9, \infty) \)[/tex]:
Choose a test point greater than 9, for example, [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = 10^2 - 12(10) + 27 = 100 - 120 + 27 = 7 \][/tex]
Since 7 is positive, [tex]\( f(x) > 0 \)[/tex] on the interval [tex]\( (9, \infty) \)[/tex].
### Step 4: Write the solution in interval notation
The polynomial [tex]\( x^2 - 12x + 27 > 0 \)[/tex] is true wherever the function is positive. Based on our analysis, these intervals are:
[tex]\[ (-\infty, 3) \cup (9, \infty) \][/tex]
### Step 5: Interpret the solution in the given table format
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline Interval & $(-\infty, 3)$ & $(3, 9)$ & $(9, \infty)$ \\ \hline Sign & positive & negative & positive \\ \hline \end{tabular} \][/tex]
Therefore, the solutions to the polynomial inequality [tex]\( x^2 - 12x + 27 > 0 \)[/tex] can be expressed in interval notation as:
[tex]\[ (-\infty, 3) \cup (9, \infty) \][/tex]
Here you also have the graph representation on the number line:
[tex]\[ \begin{array}{ccc} \text{Negative} & \text{Positive} & \text{Positive} \end{array} \][/tex]
And the intervals for which [tex]\( f(x) > 0 \)[/tex] are:
[tex]\[ x \in (-\infty, 3) \cup (9, \infty) \][/tex]
### Step 1: Set the polynomial equal to zero
Firstly, we need to find the points where the polynomial [tex]\( x^2 - 12x + 27 \)[/tex] equals zero. These points are called boundary points and can be found by solving the equation:
[tex]\[ x^2 - 12x + 27 = 0 \][/tex]
### Step 2: Solve the quadratic equation
To solve the quadratic equation, we can factorize it:
[tex]\[ x^2 - 12x + 27 = (x - 3)(x - 9) = 0 \][/tex]
Setting each factor to zero gives us the boundary points:
[tex]\[ x - 3 = 0 \implies x = 3 \][/tex]
[tex]\[ x - 9 = 0 \implies x = 9 \][/tex]
So, the boundary points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 9 \)[/tex].
### Step 3: Determine the sign intervals
The boundary points divide the number line into three intervals:
1. [tex]\( (-\infty, 3) \)[/tex]
2. [tex]\( (3, 9) \)[/tex]
3. [tex]\( (9, \infty) \)[/tex]
We need to determine whether [tex]\( x^2 - 12x + 27 \)[/tex] is positive or negative in each of these intervals.
#### Interval [tex]\( (-\infty, 3) \)[/tex]:
Choose a test point less than 3, for example, [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 12(0) + 27 = 27 \][/tex]
Since 27 is positive, [tex]\( f(x) > 0 \)[/tex] on the interval [tex]\( (-\infty, 3) \)[/tex].
#### Interval [tex]\( (3, 9) \)[/tex]:
Choose a test point between 3 and 9, for example, [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 5^2 - 12(5) + 27 = 25 - 60 + 27 = -8 \][/tex]
Since -8 is negative, [tex]\( f(x) < 0 \)[/tex] on the interval [tex]\( (3, 9) \)[/tex].
#### Interval [tex]\( (9, \infty) \)[/tex]:
Choose a test point greater than 9, for example, [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = 10^2 - 12(10) + 27 = 100 - 120 + 27 = 7 \][/tex]
Since 7 is positive, [tex]\( f(x) > 0 \)[/tex] on the interval [tex]\( (9, \infty) \)[/tex].
### Step 4: Write the solution in interval notation
The polynomial [tex]\( x^2 - 12x + 27 > 0 \)[/tex] is true wherever the function is positive. Based on our analysis, these intervals are:
[tex]\[ (-\infty, 3) \cup (9, \infty) \][/tex]
### Step 5: Interpret the solution in the given table format
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline Interval & $(-\infty, 3)$ & $(3, 9)$ & $(9, \infty)$ \\ \hline Sign & positive & negative & positive \\ \hline \end{tabular} \][/tex]
Therefore, the solutions to the polynomial inequality [tex]\( x^2 - 12x + 27 > 0 \)[/tex] can be expressed in interval notation as:
[tex]\[ (-\infty, 3) \cup (9, \infty) \][/tex]
Here you also have the graph representation on the number line:
[tex]\[ \begin{array}{ccc} \text{Negative} & \text{Positive} & \text{Positive} \end{array} \][/tex]
And the intervals for which [tex]\( f(x) > 0 \)[/tex] are:
[tex]\[ x \in (-\infty, 3) \cup (9, \infty) \][/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.