If the equation ` x^(2)+2(x+1)x+9k-5=0` has only negative roots , then :

Options

A. `k le 0`
B. `k ge 0`
C. `k ge 6`
D. `k le 6`

Answer

C

Explanation

Let `f(x) = x^(2) + 2(k-1) x + 9k - 5`. Clearly, y = f(x) represents a parabola opening upward. So, the equation f(x) = 0 will have both negative roots, if
(i) Discriminant `ge` 0 (ii) x-corrdinate of vertex `lt 0`
(iii) 0 lies outsides the roots i.e. `f(0) gt 0`
Now,
(i) Discriminant `ge 0`
`rArr" "4(k+1)^(2) - 36k + 20 ge 0`
`rArr" "k^(2) - 7k + 6 ge 0`
`rArr" "(k-1) (k-6) ge 0 rArr k le 1 or k ge 6" "...(i)`
(ii) x-coordinate of vertex `lt` 0
`rArr" "-(k+1) lt 0 rArr k+1 gt 0 rArr k gt -1" "...(ii)`
and (iii) `f(0) gt 0 rArr 9k - 5 gt 0 rArr k gt (5)/(9)" "...(iii)`
From (i), (ii) and (iii), we get `k ge 6`.