For this case we must solve the following equation of the second degree:
[tex]x ^ 2-9 = 0[/tex]
We add 9 to both sides of the equation:
[tex]x ^ 2-9 + 9 = + 9\\x ^ 2 = 9[/tex]
We apply square root on both sides of the equation:
[tex]\sqrt {x ^ 2} = \sqrt {9}\\| x | = \sqrt {9}\\x = \pm \sqrt {9}\\x = \pm3[/tex]
Answer:
[tex]x = \pm3[/tex]
The solutions are:
[tex]x=3\ and\ x=-3[/tex]
