1. ## Inequalities with radicals problem?

How should i solve this?

√x^2-9>x+6
so x^2-9<0
x<3

Im supposed to find all the possible roots or whatever and im not sure what that means/how to do it or if im even doing this problem right

2. Originally Posted by lizzytish881
How should i solve this?

√x^2-9>x+6
so x^2-9<0
x<3

Im supposed to find all the possible roots or whatever and im not sure what that means/how to do it or if im even doing this problem right
Do you understand that your posting is next to impossible to read?
Why not learn to post in symbols? You can use LaTeX tags.
$$\sqrt{x^2+1}$$ gives $\sqrt{x^2+1}$.

3. Can you solve this: SQRT(x^2 - 9) = x + 6

4. Originally Posted by Wilmer
Can you solve this: SQRT(x^2 - 9) = x + 6
I'll give you the first step. Square both sides. Then you have

${x}^{2}-9= \left( x+6 \right) ^{2}$

Now you must expand the right side. Use the FOIL method (First, outer, inner, last).

5. Arcketer, my post was addressed to the original poster...
to see if able to solve as a NON-equality...get it?

6. Yes, I suspect he understood that. The best way to solve a complicated inequality is to first solve the associated equation.

Square both sides of $\sqrt{x^2- 9}= x- 6$ to get $x^2- 18x+ 81= x^2- 12x+ 36$. Solve that for x. That divides the real numbers into two intervals. Check one value of x in each interval to see if it satisfies the inequality. If it does, then every point in the interval satisfies the inequality.

Also, in order that the square root be real, you must $x^2- 9> 0$. Again, the best way to solve that is to first solve the equation.
$x^2- 9= 0$ for x= 3 and x= -3. Checking one point of x< -3, -3< x< 3, and x> 3, we see that x< -3 and x> 3 satisfy $x^2- 9> 0$. The point in the interval above such that either of those is true satisfy the original inequality.

7. Spoiler:

the case $x\le-6$ is trivial and we easily see that the inequality holds.

in order to square both sides, we require that $x\ge-6$ as long as the radicand is well defined, putting that together we require $x\in[-6,-3].$

square and get $x<-\frac{15}4,$ thus the second solution set is $x\in \left[ -6,-\frac{15}{4} \right[$ and the final solution set is $\left] -\infty ,-6 \right]\cup \left[ -6,-\frac{15}{4} \right[=\left] -\infty ,-\frac{15}{4} \right[.$