It is given that the difference between the zeroes of

Solution:

(c) $\frac{5}{2}$

Let the zeroes of the polynomial be $\alpha$ and $\alpha+4$.

Here, $p(x)=4 x^{2}-8 k x+9$

Comparing the given polynomial with $a x^{2}+b x+c$, we get:

$a=4, b=-8 k$ and $c=9$

Now, sum of the roots $=-\frac{b}{a}$

$=>\alpha+\alpha+4=\frac{-(-8 k)}{4}$

$=>2 \alpha+4=2 k$

$=>\alpha+2=k$

$=>\alpha=(k-2) \quad \ldots(\mathrm{i})$

Also, product of the roots, $\alpha \beta=\frac{c}{a}$

$=>\alpha(\alpha+4)=\frac{9}{4}$

$=>(k-2)(k-2+4)=\frac{9}{4}$

$=>(k-2)(k+2)=\frac{9}{4}$

$=>k^{2}-4=\frac{9}{4}$

$=>4 k^{2}-16=9$

$=>4 k^{2}=25$

$=>k^{2}=\frac{25}{4}$

$=>k=\frac{5}{2} \quad(\because k>0)$