If `a, b` and `c` are positive real and `a = 2b + 3c`, then the equation `ax^(2) + bx + c = 0` has real roots for

Options

A. `|(b)/(c) - 4|ge 2sqrt(7)`
B. `|(c)/(b) - 4|ge 2sqrt(7)`
C. `|(a)/(c) - 11|ge 4sqrt(7)`
D. `|(a)/(b) + 11|ge 1sqrt((13)/(3))`

Answer

A, C

Explanation

`a = 2b + 3c`
Equation `ax^(2) + bx + c = 0` has real roots
`b^(2) - 4ac ge 0 rArr b^(2) - 4(2b + 3c)c ge 0`
`rArr c^(2)[((b)/(c))^(2) - 8((b)/(c))-12]ge0`
`rArr ((b)/(c))^(2) - 8((b)/(c)) + 16 ge 28`
`rArr ((b)/(c) - 4)^(2) ge 28 rArr |(b)/(c)-4|ge2sqrt(7)`
Similarly `|(a)/(c)-11|ge4sqrt(7)`