Question

if the roots of the equation (b-c)x²+(c-a)x+(a-b)=0are equal then prove that 2b=a+c​

Answer 1

Answer:

Step-by-step explanation

B^2-4ac=0

B^2=4AC

(C-a)^2=(4b-4c)(4a-4b)

Solving equation further we will get the answer

B=c-a,

C=a-b

Discriminate when roots are equal

D=B²-4AC=0

D=(c-a)²−4(b-c)(a-b)=0

D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0

D=c²+a²−2ac-4ab+4ac+4b²-4bc=0

c²+a²+2ac-4b(a+c)+4b²=0

(a+c)²-4b(a+c)+4b²=0

[(a+c)-2b]²=0

a+c=2b