If a,b,c are rational and a≠b,b≠a+c, then the roots of the equation (a+c−b)x2+2cx+(b+c−a)=0 are
A
irrational and distinct
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B
distinct integers
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C
rational and distinct
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D
rational and equal
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Solution
The correct option is C rational and distinct D=(2c)2−4(a+c−b)(b+c−a) ⇒D=4c2−4(c+(a−b))(c−(a−b)) =4c2−4(c2−(a−b)2) ⇒D=4(a−b)2 ⇒D>0(∵a≠b) and is a perfect square. So, roots are rational and distinct.