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|dw:1328575316576:dw|
solve for x

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

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

usually it involves the small triangle vs the large triangle

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

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

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

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

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

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

well its online so give me a sec

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

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

there is a picture of the problem!

ok upload it, I will be back in about 5 mins

ok you got my attention.

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

|dw:1328576462186:dw|

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

im sorry im just super confused!

|dw:1328576638458:dw|

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

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

yes now solve it

x=3 I used the quadratic formula

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

ok but how did you get x=3?

|dw:1328576877634:dw|

use the quadratic formula

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

Is this a challenge question?

haha ok thanks :)

thank you and again thanks for your help1