Question

please solve it and show intervals and on real line any one please explain x^2>9

Answer 1

Begin by moving the 9 over and making the inequality > 0. So, \[x ^{2}-9>0\] Now factor the left side as the difference of squares.

\[(x-3)(x+3)>0\]

Next choose values of x that are less than -3, then choose values of x that fall between -3 and 3, and finally choose values of x that are greater than 3. This will give you three intervals which you must test in your inequality to determine whether the number you chose works or not.

Let's try x=-10. If I substitute x=-10 in the factored inequality, I will get (-13)(-7) which is greater than 0, so any number less than -3 works for the inequality.

Next, let's pick a number between -3 and 3.

Let's try x=0. If I substitute x=0 in the factored inequality, I will get (-3)(3) which is not greater than 0, so any number between -3 and 3 will not work.

Finally, choose values of x which are greater than 3.

Let's try x=10. If I substitute x=10 in the factored inequality, I will get (7)(13) which is greater than 0, so any number greater than 3 works for the inequality.

So, putting the intervals together, your answer is all values of x which are in the interval below:

\[\left( -\infty,-3 \right)\cup \left( 3,\infty \right) \]