Question

If the roots of the equation (a-b)x2 + (b-c)x + (c-a) = 0 are equal, prove that 2a = b+c.

Answer 1

Answer:

Step-by-step explanation:

r + s = -(b - c)/(a - b)

but r = s:

2r = -(b - c)/(a - b)

r = -(b - c)/[2(a - b)]

also:

r * s = r^2 = (c - a)/(a - b)

(b - c)^2/[4(a - b)^2] = (c - a)/(a - b)

(b - c)^2/[4(a - b)] = (c - a)

b^2 - 2bc + c^2 = 4ac - 4a^2 - 4bc + 4ab

4a^2 - 4ab + b^2 - 4ac + 2bc + c^2 = 0

(2a - b)^2 - 2c(2a - b) + c^2 = 0

[(2a - b) - c]^2 = 0

2a - b - c = 0

2a = b + c

Answer 2

The question must be, if roots of equation (a-b)x2+(b-c)x+(c-a)=0 are equal , prove that b+c = 2a.

Using Discriminant,

D = B2-4AC as compared with the general quadratic equation Ax2+Bx+C=0

so, A = a-b

B = b-c

C = c-a

For roots to be equal, D=0

(b-c)2 - 4(a-b)(c-a) =0

b2+c2-2bc -4(ac-a2-bc+ab) =0

b2+c2-2bc -4ac+4a2+4bc-4ab=0

4a2+b2+c2+2bc-4ab-4ac=0

(2a-b-c)2=0

i.e. 2a-b-c =0

2a= b+c

Hope this will help you...!!