If roots α, β of the equation x

Given equation: $x^{2}-p x+16=0$

Also, $\alpha$ and $\beta$ are the roots of the equation satisfying $\alpha^{2}+\beta^{2}=9$.

From the equation. we have:

Sum of the roots $=\alpha+\beta=-\left(\frac{-p}{1}\right)=p$

Product of the roots $=\alpha \beta=\frac{16}{1}=16$

Now, $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$

$\Rightarrow p^{2}=9+32$

$\Rightarrow p^{2}=41$

$\Rightarrow p=\sqrt{41}$

Hence, the value of $p$ is $\sqrt{41}$.