Another year of school, another first couple weeks of review. I come to the pythagrean theorum: a^2 + b^2 = c^2

I understand how it works, so I guess what I am asking for is algebra help. I haven't done math all summer, so this could be a long time before everything starts clicking again. Okay,

a is (x-4)
b is (x-3)
c is (12-x)

I get it too: (x-4)^2 + (x-3)^2 = (12-x)^2

Now I don't know what steps to do in order to solve for x. Could someone PLEASE walk me through it? I'll go look in my textbook to see what it says the answer is, just so we're all sure.

EDIT:
Okay, the book says the answer is 7.
Last edited by pepsi_lovr at Sep 8, 2008,
just multiply everything and add it together and put it all on one side

(x-4)*(x-4) = x^2-8x+16
(x-3)*(x-3) = x^2-6x+9
(12-x)*(12-x) = 144-24x+x^2

x^2-8x+16 + x^2-6x+9 = 144-24x+x^2
x^2 + 10x - 119 = 0

Quote by seljer
just multiply everything and add it together and put it all on one side

(x-4)*(x-4) = x^2-8x+16
(x-3)*(x-3) = x^2-6x+9
(12-x)*(12-x) = 144-24x+x^2

x^2-8x+16 + x^2-6x+9 = 144-24x+x^2
x^2 + 10x - 119 = 0

this
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Quote by seljer
just multiply everything and add it together and put it all on one side

(x-4)*(x-4) = x^2-8x+16
(x-3)*(x-3) = x^2-6x+9
(12-x)*(12-x) = 144-24x+x^2

x^2-8x+16 + x^2-6x+9 = 144-24x+x^2
x^2 + 10x - 119 = 0

Thank you. I get an answer of 7.08, I assume that this would count as correct as it is only 0.08 off of the answer given in the book.

You make me a happy man.
Quote by pepsi_lovr
Thank you. I get an answer of 7.08, I assume that this would count as correct as it is only 0.08 off of the answer given in the book.

You make me a happy man.

you did something wrong

the answers are precisely 7 and -17 (though if you're talking about the sides of a triangle here you can cross that one off as we don't really have negative distances)

in this case
a= 1
b=10
c=-119