Question

If a, b, c ∈ R and x² + (a + b)x + c = 0 has no real roots, then

(A)  c (a + b + c ) > 0

(B) c + c (a + b + c ) > 0 

(C) c + c (a + b − c ) > 0

(D) c (a + b − c ) > 0

Answer 1

Correct option (C) c + c (a + b − c ) > 0

Since the roots of ax² + bx + c = 0 are non-real, thus f(x) = ax² + bx + c will have same sign for every value of x.

f(0) = c 

f(1) = a + b + c 

f(−1) = a − b + c 

f(2) = 4a − 2b + c

⇒ c⋅(a + b + c) > 0, c(a − b + c) > 0

⇒ c (4a − 2b + c) > 0