Respuesta :

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]

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