The roots of ` ax^(2) +bx +c =0 " whose " a ne 0, b ,c in R `, " are non-real complex and " a + c lt b, " then

Options

A. `4a + c gt 2b`
B. `4a + c lt 2b`
C. `4a + c = 2b`
D. none of these

Answer

B

Explanation

It is given that `ax^(2) + bx + c = 0` has complex roots. Then,
`b^(2) - 4ac lt 0 rArr a and c` are of the same sign. ltBrgt Now, two case arise.
CASE I When a and c are both positive
In this case, we have `a+c lt b`
`rArr" "(a+c)^(2) lt b^(2)`
`rArr" "(a + c)^(2) lt 4ac" "[because b^(2) lt 4ac]`
`rArr" "(a-c)^(2) lt 0` which is not possible.
CASE II When a and c are both negative
In this case, we have `a + c lt b and b^(2) - 4ac lt 0`
Clearly, when a and c are both negative, b must be positive. Otherwise, the equation becomes one where coefficients are all positive. Therefore, `4a + c lt 2b`.