Correct Answer - A
Let `p(x) =x^(2)+kx +k, k ne 0`
On comparing `p(x)` with `ax^(2)+bx +c`, we get
`a = 1, b = k` and `c = k`
Now, `x = (-b +-sqrt(b^(2)-4ac))/(2a)` [by quadratic formula]
`= (-k+- sqrt(k^(2)-4k))/(2xx1)`
`= (-k +- sqrt(k(k-4)))/(2), k ne 0`
Here, we see that
`k(k-4) gt 0`
`rArr k in (-oo,0) uu (4,oo)`
Now, we know that
In quadratic polynomial `ax^(2) +bx +c`
If `a gt 0, b gt 0, c gt 0` or `a lt 0, b lt 0, c lt 0`,
then the polynomial has always all negative zeroes. and is `a gt 0, c lt 0` or `a lt 0, c gt 0`, then the polynomial has always zeroes of opposite sign.
Case I if `k in (-oo,0)` i.e., `k lt 0`
`rArr a = 1 gt 0, b,c = k lt 0`
So, the zeroes are of opposite sign.
Case II If `k in (4, oo) i.e., k ge 4`
`rArr a = 1 gt 0, b ge 4`
So, both zeroes are negative
Hence, in any case zeroes of the given quadratic polynomial cannot both be positive.
