For what value of a is there a right triangle with sides a+1, 6a, and 2a+1?
Solve for a:
36 a^2 = (a + 1)^2 + (2 a + 1)^2
Expand out terms of the right hand side:
36 a^2 = 5 a^2 + 6 a + 2
Subtract 5 a^2 + 6 a + 2 from both sides:
31 a^2 - 6 a - 2 = 0
Divide both sides by 31:
a^2 - (6 a)/31 - 2/31 = 0
Add 2/31 to both sides:
a^2 - (6 a)/31 = 2/31
Add 9/961 to both sides:
a^2 - (6 a)/31 + 9/961 = 71/961
Write the left hand side as a square:
(a - 3/31)^2 = 71/961
Take the square root of both sides:
a - 3/31 = sqrt(71)/31 or a - 3/31 = -sqrt(71)/31
Add 3/31 to both sides:
a = 3/31 + sqrt(71)/31 or a - 3/31 = -sqrt(71)/31
Add 3/31 to both sides:
a = 3/31 + sqrt(71)/31 ==0.36859