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प्रश्न
If the roots of the equation (b − c) x2 + (c − a) x + (a − b) = 0 are equal, then prove that 2b = a + c.
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उत्तर
The given quadric equation is (b − c) x2 + (c − a) x + (a − b) = 0, and roots are real
Then prove that 2b = a + c
Here,
a = (b − c), b = (c − a) and c = (a − b)
As we know that D = b2 − 4ac
Putting the value of a = (b − c), b = (c − a) and c = (a − b)
D = b2 − 4ac
= (c − a)2 − 4 × (b − c) × (a − b)
= c2 − 2ca + a2 − 4 (ab − b2 − ca + bc)
= c2 − 2ca + a2 − 4ab + 4b2 + 4ca − 4bc
= c2 + 2ca + a2 − 4ab + 4b2 − 4bc
= a2 + 4b2 + c2 + 2ca − 4ab − 4bc
As we know that (a2 + 4b2 + c2 + 2ca − 4ab − 4bc) = (a + c − 2b)2
D = (a + c − 2b)2
The given equation will have real roots, if D = 0
(a + c − 2b)2 = 0
Square root both side we get
`sqrt((a + c - 2b)^2)=0`
a + c − 2b = 0
a + c = 2b
Hence 2b = a + c.
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