## anonymous 5 years ago i need help with the following attachment ....

1. anonymous

2. anonymous

What do you need help with specifically. Long problem, has a lot of work on it already?

3. anonymous

to check #6 if i did it correctlty?

4. anonymous

Good work, unfortunately some algebra mistakes. When working on an equation, If you do something to one side, you have to do to the other side (all terms). Likewise when you do something to a radical expression p/q; you must do top and bottom.

5. anonymous

ahh i see so then that would make it x = 400*666.7?

6. anonymous

On your first step you square, square root (400 + x^2) and 1000 but you didn't square 1500. In any event, it is not recommended to go this route. When you want to square to remove radical, you should first isolate the radical to one side and all other terms to the other side. So your first step, it is recommended to multiply through by sq rt (400+x^2) not to remove radical but to cancel it from the denominator.

7. anonymous

okay give me a sec and ill do it

8. anonymous

so x=(2/3)?

9. anonymous

That was a big jump from one step to the next, may be it's right but the next step is $-1000(\sqrt{400+x ^{2}}) +1500x=0$

10. anonymous

aww no im not right i did something else..okay let me see if i can get it

11. anonymous

is the next step moving the "1000" over the "=" sign making it a positive?

12. anonymous

You can, but it is just as easy to divide by -1000.

13. anonymous

so it just basically cancels out?

14. anonymous

$\sqrt{400+x ^{2}}=(-1500x)/(-1000)$

15. anonymous

wouldnt it be positive 1000?

16. anonymous

oh wait nevermind...

17. anonymous

would it be -1500x=$\sqrt{400+x ^{2}}$(-1000) ?

18. anonymous

The purpose of doing this is to isolate the radical, then square it and get rid of it$400+x ^{2}=(9x ^{2})/4$

19. anonymous

how did you get $(9x ^{2})/4??$

20. anonymous

$[(-1500x)/(-1000)]^{2}=(15x/10)^{2}=(3x/2)^{2}$

21. anonymous

does x= -31???

22. anonymous

I am not confident in your work, you jump too many steps; it's not the way you do mathematics. Work it from there. I'll check it later. It is definitely not -31

23. anonymous

okay i went from your step to $x^{2}=(9^{x ^{2}}-400]?? 24. anonymous \[x ^{2}=(9x ^{2}-400)$

25. anonymous

9xsquared /4 that is sorry

26. anonymous

i got $100\sqrt{2}$

27. anonymous

$x ^{2}-(9x ^{2})/4=-400$

28. anonymous

$x ^{2}(1-[9/4]=-400$

29. anonymous

$x ^{2}(-5/4)=-400$

30. anonymous

31. anonymous

$x ^{2}=400(4/5)$

32. anonymous

now we divide -400 and (-5/4)?

33. anonymous

Yes$x =\sqrt{320}$$x =6\sqrt{5}$

34. anonymous

ahhh i had $\sqrt{200}$ .... boy was i off ... well thank you so much for helping me i have a better understanding for calculus im in 10th grade so im new to this ... you were tons of help ..bye bye?!

35. anonymous

Good work, you are doing good. It took years of working with a tutor for me to feel this comfortable.