Question:
If the difference of the roots of the equation x2 – Px + 8 = 0 is 2, then P =___________.
Solution:
For $x^{2}-P x+8=0$
Let us suppose two roots are $z_{1}$ and $z_{2}$
given : $z_{1}-z_{2}=2$
also $z_{1} z_{2}=8$n (product of roots)
i. e. $\left(z_{2}+2\right) Z_{2}=8$
i. e. $z_{2}^{2}+2 z_{2}=8$
i.e. $z_{2}^{2}+2 z_{2}-8=0$
i. e. $z_{2}^{2}+4 z_{2}-2 z_{2}-8=0$
i. e. $\left(z_{2}+4\right)\left(z_{2}-2\right)=0$
i.e. $z_{2}=-4$ or $z_{2}=2$
$\Rightarrow z_{1}=-2$ or $z_{2}=4$
Since $P=z_{1}+z_{2}$
$\Rightarrow P=-4-2$ or $P=4+2$
i.e. $P=-6$ or 6