Solve the inequality and express your answer in interval notation.
x2 - 12x + 5 <0

Answer:
([tex]6 - \sqrt{31}[/tex], [tex]6 + \sqrt{31}[/tex])
Step-by-step explanation:
[tex]x^{2}-12x+5 < 0[/tex]
To solve this inequality, we can start by completing the square. To do this, we divide the x coefficient (-12) by 2 and then square it. We then add the result to the left side of the equation. To ensure the inequality remains true, we also subtract it.
[tex]x^{2}-12x+(-6^{2})-(-6^{2}) + 5 < 0\\x^{2}-12x+36-36 + 5 < 0\\[/tex]
We can then express the first trinomial as a perfect square.
[tex](x-6)^{2} - 31 < 0[/tex]
Add 31 to both sides:
[tex](x-6)^{2}< 31[/tex]
Then take the square root of both sides:
[tex]x-6<[/tex] ± [tex]\sqrt{31}[/tex]
Finally, add 6 to both sides:
x < 6 + [tex]\sqrt{31}[/tex]
x > 6 - [tex]\sqrt{31}[/tex]