The two legs are 2x-1 and x
The hypotenuse is 2x+1
What is x?
2007-05-10 13:38:52 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(2x-1)^2 + x^2 = (2x+1)^2
4x^2-4x+1+x^2 = 4x^2+4x+1
x^2 - 8x = 0
x(x-8) = 0
x = 0, 8
but you can't use 0, because a leg of a triangle can't be 0, so the answer is x=8.
2007-05-10 13:43:05 · answer #1 · answered by David 3 · 0⤊ 0⤋
a^2 + b^2 = c^2 is the formula for the relationship between the sides a and b and the hypotenuse c of a right angled triangle.
(2x - 1)^2 + x^2 = (2x + 1)^2
4x^2 - 4x + 1 + x^2 = 4x^2 + 4x + 1
Transferring the terms on the right hand side to the Left, we can write
x^2 - 8x = 0
x ( x - 8 ) = 0
So, we have two solutions for x, x = 0 and x = 8
Out of these two, x = 0 is an imaginary solution since 2x - 1 will be -1 which is not really possible. So, we accept x = 8 as the solution and we cross check.
15^2 + 8^2 = 225 + 64 = 289 = 17^2
So, x = 8
2007-05-10 20:52:42 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
(2x+1)^2 = (2x-1)^2 + x^2
4x^2 + 4x + 1 = 4x^2 - 4x + 1 + x^2
x^2 -8x = 0
x(x-8) = 0
x = 0, x = 8
8 only
Check:15, 8, 17 is a right triangle
2007-05-10 20:44:25 · answer #3 · answered by richardwptljc 6 · 0⤊ 0⤋
(2X+1)^2=(2X-1)^2 + X^2
4X^2 + 4X + 1 = 4X^2- 4X +1 +X^2
8X= X^2
X=8
2007-05-10 20:47:50 · answer #4 · answered by nick w 1 · 0⤊ 0⤋
(L1)^2 + (L2)^2 = H^2...rest is just applying the formula
2007-05-10 20:49:12 · answer #5 · answered by Bladvak 3 · 0⤊ 0⤋