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If a, b and c are the rational numbers (a > b > c > 0) and the quadratic equation (a + b − 2c)x2 + (b + c − 2a)x + (c + a − 2b) = 0 has a root in the interval (−1, 0), then

(A)  c + a < 2b.

(B) Both roots are rational.

(C) The equation ax2 + 2bx + c = 0 has both negative real roots.

(D) The equation cx2 + 2ax + b = 0 has both negative real roots.

1 Answer

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Correct option (A), (B), (C), (D)

f(−1) f(0) < 0

⇒ (2a − b − c) (c + a − 2b) < 0

⇒ (a − b + a − c) (c + a − 2b) < 0

⇒ c + a < 2b

One root is 1 and other is c + a - 2b/a + b - 2c

Therefore, both roots are rational.

Now, discriminant of ax2 + 2bx + c = 0 is 4b2 − 4ac.

Using (c + a) < 2b, we have D > 0.

Also, a, b and c are +ve. Therefore, both the roots are real and −ve.

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