The zeroes of the quadratic polynomial `x^(2) +kx +k` where `k ne 0`, are : (1)cannot both be positive (2)cannot both be negative (3)are always unequal (4)are always equal

Options

A. cannot both be positive
B. cannot both be negative
C. are always unequal
D. are always equal

Answer

A

Explanation

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.