Answer the following and justify:
Can the quadratic polynomial $x^{2}+kx+k$ have equal zeroes for some odd integer $k > 1$?

(A) No. Let $p(x) = x^{2} + kx + k$. If $p(x)$ has equal zeroes,its discriminant $D$ must be equal to $0$.
The discriminant formula is $D = B^{2} - 4AC = 0$ ... $(i)$.
Comparing $p(x)$ with the standard form $Ax^{2} + Bx + C$,we get $A = 1$,$B = k$,and $C = k$.
Substituting these values into Eq. $(i)$:
$k^{2} - 4(1)(k) = 0$
$k^{2} - 4k = 0$
$k(k - 4) = 0$
This gives $k = 0$ or $k = 4$.
Since neither $0$ nor $4$ is an odd integer greater than $1$,the quadratic polynomial $x^{2} + kx + k$ cannot have equal zeroes for any odd integer $k > 1$.