Question:
If roots $\alpha, \beta$ of the equation $x^{2}-p x+16=0$ satisfy the relation $\alpha^{2}+\beta^{2}=9$, then write the value $p$.
Solution:
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 $\sqrt{41}$.