Let `f(x) =ax^2 + bx+ c AA a, b, c in R, a != 0` satisfying `f (1) + f(2)=0`. Then, the quadratic equation `f(x)=0` must have :

Options

A. no real root
B. 1 and 2 as real roots
C. two equal roots
D. two distinct real roots

Answer

D

Explanation

We have, `f(x) = ax^(2) + bx + c`
`therefore" "f(1) + f(2) = 0 rArr 5a + 3b + 2c - 0`
Let D be the discriminant of f(x) = 0. Then,
`D = b^(2) - 4ac`
`rArr" "D=((5a+2c)/(-3))^(2)-4ac" "["Using (i)"]`
`rArr" "D=((5a+2c)^(2)-36ac)/(9)`
`rArr" "D=(25a^(2)+4c^(2)-16ac)/(9)=(1)/(9){(4a-2c)^(2)+9a^(2)} gt 0`
Hence, f(x) = 0 has two distinct real roots.