Show that the roots r, s, t of the equation x(x - 2)(3x - 7) = 2 are real and positive. Find tan-1r + tan-1s + tan-1t.
Solution
Put f(x) = x(x - 2)(3x - 7) - 2 = 3x3 - 13x2 + 14x - 2. Then f(0) = -2, f(1) = 2, so there is a root between 0 and 1. f(2) = -2, so there is another root between 1 and 2. f(3) = 4, so the third root is between 2 and 3. f(x) = 0 has three roots, so they are all real and positive.
We have tan(a + b + c) = (tan a + tan b + tan c - tan a tan b tan c)/(1 - (tan a tan b + tan b tan c + tan c tan a)). So putting a = tan-1r, b = tan-1s, c = tan-1t, we have, tan(a + b + c) = ( (r + s + t) - rst)/(1 - (rs + st + tr) ) = (13/3 - 2/3)/(1 - 14/3) = -1. So a + b + c = -π/4 + kπ. But we know that each of r, s, t is real and positive, so a + b + c lies in the range 0 to 3π/2. Hence a + b + c = 3π/4.
© John Scholes
jscholes@kalva.demon.co.uk
1 July 2002