Let a be a real number for which there exists a unique value of b such that the quadratic equation x2+2bx+(a−b)=0 has one real solution. Find a .
one real solution means that the discriminant is zeroin the quadratic equation ax2+bx+c the discriminant is given by D=b2−4acherea=1, b=2b, c=a−bD=(2b)2−4(a−b)=4b2−4a+4ab D=0⇒4b2−4a+4ab=0a=b+b2
To have one real solution the discriminant must = 0 ....so...
(2b)^2 - 4 (1) (a - b) = 0
4b^2 - 4a + 4b = 0
b^2 - a + b = 0
b^2 + b = a
