If a,b,c are distinct non-zero rational numbers such that a+b+c=0, then both the roots of the equation (b+c−a)x2+(c+a−b)x+(a+b−c)=0 are
A
rational and unequal
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B
irrational
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C
non-real
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D
equal
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Solution
The correct option is A rational and unequal (b+c−a)x2+(c+a−b)x+(a+b−c)=0
As a+b+c=0, the equation becomes −2ax2−2bx−2c=0 ⇒ax2+bx+c=0
Putting x=1 a+b+c=0
So, x=1 is a root of the equation.
Let the other root be α
Product of roots, 1×α=ca⇒α=ca
As a and c both are rational numbers, α is also rational.