## anonymous 5 years ago how would you solve (sqrt(x+2)) + (sqrt(3-x))

1. anonymous

$\sqrt{x+2}+\sqrt{3-x}=0 ?$

2. anonymous

$\sqrt{a} ~~~\,a \ge0$

3. anonymous

$x+2\ge0 \,and \,~3-x \ge0$

4. anonymous

f(x) = $\sqrt{x+2}$ ; g(x) = $\sqrt{3-x}$ find (f+g)(x) how would I do that?

5. anonymous

$x \ge2~\, and~\,x \le3$

6. anonymous

ohh ok

7. anonymous

(f+g)(x)=f(x)+g(x) $(f+g)(x)=\sqrt{x+2}+\sqrt{3-x}$ this is the answer you cant do anything

8. anonymous

what if it was (f * g) ? could you do something then?

9. anonymous

since inside of the roots are different you cannot make addition

10. anonymous

is it multiply sign?

11. anonymous

yeah

12. anonymous

if it was (f*g)(x), you can multiply the expression inside the radical signs since they have common indices.

13. anonymous

(f*g)(x)=f(x)*g(x) $(f*g)(x)=(\sqrt{x+2})(\sqrt{3-x})$ $\sqrt{(x+2)(3-x)}$$\sqrt{-x^2+x-6}$

14. anonymous

it should be +6 not -6

15. anonymous

o okay that makes sense thank you

16. anonymous

yw