Answer :
To determine the solution for the inequality [tex]\(x^2 + 36 > 12x\)[/tex], let's go through a detailed, step-by-step process of solving it.
1. Rearrange the inequality:
Start by rearranging the inequality to bring all terms to one side of the inequality.
[tex]\[ x^2 + 36 > 12x \][/tex]
Subtract [tex]\(12x\)[/tex] from both sides:
[tex]\[ x^2 - 12x + 36 > 0 \][/tex]
2. Factorize the quadratic expression:
We need to recognize that the quadratic expression [tex]\(x^2 - 12x + 36\)[/tex] can be factorized. Notice that it represents a perfect square:
[tex]\[ x^2 - 12x + 36 = (x - 6)^2 \][/tex]
So, the inequality can be rewritten as:
[tex]\[ (x - 6)^2 > 0 \][/tex]
3. Analyze the inequality:
A squared term [tex]\((x - 6)^2\)[/tex] is always non-negative (greater than or equal to 0). The term [tex]\((x - 6)^2\)[/tex] is equal to 0 when [tex]\(x = 6\)[/tex]. We are interested in when it is strictly greater than 0.
[tex]\[ (x - 6)^2 > 0 \quad \text{when} \quad x \neq 6 \][/tex]
4. Solution set:
The inequality [tex]\((x - 6)^2 > 0\)[/tex] is true for all real numbers [tex]\(x\)[/tex] except [tex]\(x = 6\)[/tex].
Therefore, the solution to the inequality [tex]\(x^2 + 36 > 12x\)[/tex] is:
[tex]\[ \{x \mid x \in \mathbb{R} \text{ and } x \neq 6\} \][/tex]
Thus, the correct solution is:
[tex]\[ \boxed{\{ x \mid x \in \mathbb{R} \text{ and } x \neq 6 \}} \][/tex]
1. Rearrange the inequality:
Start by rearranging the inequality to bring all terms to one side of the inequality.
[tex]\[ x^2 + 36 > 12x \][/tex]
Subtract [tex]\(12x\)[/tex] from both sides:
[tex]\[ x^2 - 12x + 36 > 0 \][/tex]
2. Factorize the quadratic expression:
We need to recognize that the quadratic expression [tex]\(x^2 - 12x + 36\)[/tex] can be factorized. Notice that it represents a perfect square:
[tex]\[ x^2 - 12x + 36 = (x - 6)^2 \][/tex]
So, the inequality can be rewritten as:
[tex]\[ (x - 6)^2 > 0 \][/tex]
3. Analyze the inequality:
A squared term [tex]\((x - 6)^2\)[/tex] is always non-negative (greater than or equal to 0). The term [tex]\((x - 6)^2\)[/tex] is equal to 0 when [tex]\(x = 6\)[/tex]. We are interested in when it is strictly greater than 0.
[tex]\[ (x - 6)^2 > 0 \quad \text{when} \quad x \neq 6 \][/tex]
4. Solution set:
The inequality [tex]\((x - 6)^2 > 0\)[/tex] is true for all real numbers [tex]\(x\)[/tex] except [tex]\(x = 6\)[/tex].
Therefore, the solution to the inequality [tex]\(x^2 + 36 > 12x\)[/tex] is:
[tex]\[ \{x \mid x \in \mathbb{R} \text{ and } x \neq 6\} \][/tex]
Thus, the correct solution is:
[tex]\[ \boxed{\{ x \mid x \in \mathbb{R} \text{ and } x \neq 6 \}} \][/tex]
