Question

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Answer 1

Answer:

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

Step-by-step explanation:

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Answer 2

Answer:

Since, roots are equal.

so,

b²-4ac=0

-> (b-c)²-4×(c-a)×(a-b)=0

-> b²+c²-2bc-4(ac-bc-a²+ab)

-> b²+c²-2bc-4ac+4bc+4a²-4ab=0

-> 4a²+b²+c²-4ac+2bc-4ab=0

(a+b+c)²=a²+b²+2ab+2bc+2ac

-> (-2a+b+c)²=0

-> -2a+b+c=0

-> b+c=2a