Question

If the roots of the equation
(b-c)x²+(a-c)x+(a-b)=0 are equal then prove that 2b=a+c.
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Answer 1

Answer:

Given,

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

The roots of the given equation are equal

therefore, D = b^2 - 4ac =0

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

=>(c^2 + a^2 - 2ac) - 4(ab - b^2 - ca + bc) = 0

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

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

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

=>(a + (-2b) + c)^2 =0. {(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx}

=>a - 2b + c = 0

=>2b = a+c

Answer 2

Answer:

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