For two real values of n, the equation 4x^2+nx+25 has exactly one solution in x. What is the positive value of n?
there are only a few ways to setup this into factored form
(2x+-25)(2x+-1)
(4x+-1)(x+-25)
(4x+-25)(x+-1)
(4x+-5)(x+-5)
(2x+-5)(2x+-5)
if we want sqrt of B^2-4ac to have one solution that has to equal 0
the only one where B^2 = 400 is when b = 20 or (2x+-5)(2x+-5)
so N or b = 20
