## anonymous 4 years ago Please help... so confused HOLD ON! im going to draw a picture!

1. anonymous

|dw:1328575316576:dw| solve for x

2. precal

use \[a^2+b^2=c^2\]

3. anonymous

well i think that would work but we're supposed to make it a proportion

4. precal

usually it involves the small triangle vs the large triangle

5. anonymous

ya but i cant do that... at least i dont think i can

6. precal

do you know about the angle on top? Is is split in the middle? Is it 45 degrees since the larger part is 90 degrees? I can't tell by your drawing. I know it is hard to draw

7. anonymous

well the whole triangle is 90 degrees but it doesnt tell you if its split through the middle

8. precal

|dw:1328575806467:dw| if so we can assume the side that is in between them is x...

9. anonymous

but how can you do that since x is the length of the side on the bottom?

10. precal

Is the drawing from a book or a worksheet? the problem with geometry is that you can not assume anything

11. anonymous

its in the book and ya thats why i dont like it!

12. precal

|dw:1328575902416:dw| We can assume that 90 degrees, the other one we can assume is the 45 degrees because that line that splits the triangle has a special name but I can't recall it at the moment.

13. precal

You can scan the picture and attach the file. but I think I might have it figured out

14. anonymous

well its online so give me a sec

15. precal

\[x^2+x^2=(x+3)^2\] \[2x^2=x^2+6x+9\] \[x^2-6x-9=0\] not that does not work !!!!!

16. precal

I meant that does not work!!! Sorry can't seem to type as well

17. anonymous

there is a picture of the problem!

18. precal

19. precal

ok you got my attention.

20. precal

ok x+3 is the leg of the big triangle the hypotenuse is 12-x for the big triangle

21. anonymous

well the problem never tells you anything about the angles or anything and as you can tell the line doesnt go right through the right angle symbol

22. precal

ok x is the leg of the small triangle and x+3 is the hypotenuse of the small triangle there is our proportion

23. precal

|dw:1328576462186:dw|

24. anonymous

but how did you get 12-x since 12 is the whole LARGE triangle's hypotenuse?

25. anonymous

im sorry im just super confused!

26. precal

|dw:1328576638458:dw|

27. anonymous

so (x+3)(x+3)=12 x

28. precal

No, I think we are ok cross multiply \[12x=(x+3)^2\]

29. precal

yes now solve it

30. anonymous

x^2+3x+3x+9=12x x^2+6x+9=12x x^2+9=6x right? but i dont know what do to next because we havent done anything like this in a while!

31. precal

x=3 I used the quadratic formula

32. precal

\[x^2-6x+9=0\]

33. anonymous

ok but how did you get x=3?

34. precal

|dw:1328576877634:dw|

35. precal

36. anonymous

see all of that confuses me and i dont think we've done that yet!

37. anonymous

well thanks for your help but i dont think im going to get it.... but thank you for your trying... :)

38. precal

Is this a challenge question?

39. anonymous

i dont think so.... but online they sometimes leave off something like a picture or a number so that might be what happened... thanks for you patients.

40. precal

You will understand this eventually, your teacher has to teach quadratics. One day it will make sense.

41. anonymous

haha ok thanks :)

42. precal

Oh well another reason why computers will not replace teachers. I am sure that it is an oversight of the program. Hang in there :)

43. anonymous

thank you and again thanks for your help1