Find all values of `p` so that 6 lies between roots of the equation `x^(2)+2(p-3)x+9=0`

Options

A. `a in [-3//4, oo)`
B. `a in (oo, -3//4)`
C. `a in (-oo, 0) uu(6, oo)`
D. `a in (-3//4, 6)`

Answer

B

Explanation

Let `f(x) = x^(2) + 2(a-3) x + 9`.
If 6 lies between the roots fo f(x) = 0, then we must have
(i) `"Disc" gt 0, and" "(ii) f(6) lt 0" "[therefore "Coeff of" x^(2) "is positive"]`
Now, Disc `gt` 0
`rArr" "4(a-3)^(2) - 36 gt 0`
`rArr" "(a-3)^(2) - 9 gt 0`
`rArr" "a^(2) - 6a gt 0`
`rArr" "a(a-6) gt 0 rArr a lt 0 or a gt 6" "...(i)`
and, `f(6) lt 0`
`rArr" "36 + 12(a-3) + 9 lt 0 rArr a lt -(3)/(4)" "....(ii)`
From (i) and (ii), we get
`a lt - 3//4 i.e. a in (-oo, -3//4)`.