Question

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

Answer 1

Answer:

($6 - \sqrt{31}$, $6 + \sqrt{31}$)

Step-by-step explanation:

$x^{2}-12x+5 < 0$

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.

$x^{2}-12x+(-6^{2})-(-6^{2}) + 5 < 0\\x^{2}-12x+36-36 + 5 < 0$

We can then express the first trinomial as a perfect square.

$(x-6)^{2} - 31 < 0$

Add 31 to both sides:

$(x-6)^{2}< 31$

Then take the square root of both sides:

$x-6<$ ± $\sqrt{31}$

Finally, add 6 to both sides:

x < 6 + $\sqrt{31}$

x > 6 - $\sqrt{31}$