Question

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

Answer 1

     

The roots are equal, so the discriminant is 0.

$\Rightarrow\  (b - c)^2 - (4(a - b)(c - a)) = 0 \\ \\ \Rightarrow\  (b^2 - 2bc + c^2) - (4(ac - a^2 - bc + ab)) = 0 \\ \\ \Rightarrow\  b^2 - 2bc + c^2 - (4ac - 4a^2 - 4bc + 4ab) = 0 \\ \\ \Rightarrow\  b^2 - 2bc + c^2 - 4ac + 4a^2 + 4bc - 4ab = 0\\ \\ \Rightarrow\  4a^2 + b^2 + c^2 - 4ab + 2bc - 4ac = 0 \\ \\ \Rightarrow\  (2a - b - c)^2 = 0 \\ \\ \Rightarrow\  2a - b - c = 0 \\ \\ \Rightarrow\  2a = b + c$

Hence proved!

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Answer 2

Answer:

Step-by-step explanation:

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