Asked by Aakash
If the roots of the equation (b-c)x^2+(c-a)x+(a-b)=0 are equal, prove that 2b=a+c.
Answers
Steve
The discriminant must be zero, so
(c-a)^2-4(b-c)(a-b)=0
c^2+a^2−2ac−4(ab−b^2−ac+bc)=0
c^2+a^2−2ac−4ab+4b^2+4ac−4bc=0
c^2+a^2+2ac+4b^2−4ab−4bc=0
(c+a)^2−4b(a+c)+4b^2=0
(c+a)^2−2⋅(c+a)⋅(2b)+(2b)^2=0
[(c+a)−(2b)]^2=0
c+a−2b=0
2b = a+c
(c-a)^2-4(b-c)(a-b)=0
c^2+a^2−2ac−4(ab−b^2−ac+bc)=0
c^2+a^2−2ac−4ab+4b^2+4ac−4bc=0
c^2+a^2+2ac+4b^2−4ab−4bc=0
(c+a)^2−4b(a+c)+4b^2=0
(c+a)^2−2⋅(c+a)⋅(2b)+(2b)^2=0
[(c+a)−(2b)]^2=0
c+a−2b=0
2b = a+c