Geometry Question

[fatty4ksu]

Originally posted by: PurdueRy

Originally posted by: chuckywang
PM me for physics, Math, or EE help

OMG. Frikin LOL

 

[BoldAsLove]

i made a mistake...its 4x+1...does that make it easier?

 

[chuckywang]

Originally posted by: PurdueRy

Originally posted by: chuckywang
Did it in my head. x=160/7

fixed

GOSHDARNIT!!

 

[bonkers325]

yes, x = 7

 

[cheezmunky]

Originally posted by: cheezmunky
this isn't a right triangle and it isn't solvable without an angle

nevermind I thought x couldn't be negative but it can

(4x + 20)^2 = 20^2 + (3x)^2
16x + 160x + 400 = 400 + 9x^2
7x + 160x = 0

x = 0 , -(160/7)

 

[JujuFish]

Originally posted by: BoldAsLove
i made a mistake...its 4x+1...does that make it easier?

That makes x = 7.

 

[BoldAsLove]

DOH! thats what i originally had!

 

[ArJuN]

Originally posted by: BoldAsLove
i made a mistake...its 4x+1...does that make it easier?

god damn, yes it does.

a^2 + b^2 = c^2
a = 3x
b = 20
c = 4x + 1
(3x)^2 + (20)^2 = (4x+1)^2
9X^2 + 400 = 16X^2 + 4x + 4x + 1
9X^2 + 400 = 16X^2 + 8x + 1
0 = 7x^2 + 8x - 399
now solve for x

EDIT:it's 7 i believe.

 

[cheezmunky]

Originally posted by: cheezmunky

Originally posted by: cheezmunky
this isn't a right triangle and it isn't solvable without an angle

nevermind I thought x couldn't be negative but it can

(4x + 20)^2 = 20^2 + (3x)^2
16x + 160x + 400 = 400 + 9x^2
7x^2 + 160x = 0

x = 0 , -(160/7)

I take that back once again. that yields negative values for the lengths of the sides, making my original statement true