p5^y + 5^-y = 5
p5^y + 1/5^y = 5
now replace
px + 1/x = 5
times x
px^2 + 1 = 5x
px^2 - 5x + 1 = 0
which is what we want
By letting x=5^y, show that the equation p5^y + 5^-y = 5 can also be written as px^2-5x+1 = 0
1 answer