#1**+2 **

"the polynomial x^2+bx+c has exactly one real root" => Discriminant : b^2 - 4*a*c = 0

but a = 1 => b^2-4c = 0 (1)

we also have : b = c+1 (2)

Solve for the set of equation of (1) and (2) => c = 1.

noobfromvn26 Feb 9, 2020

#2**+2 **

About q2:

x^2 - 3x + 1 = x^2 - 2 * (3/2) * x + (9/4) - (9/4) + 1 = (x - 3/2)^2 - 5/4.

=> b = -3/2 ; c = -5/4

noobfromvn26 Feb 9, 2020