Question

If the difference of the roots of the equation x² – Px + 8 = 0 is 2, then P =___________.

Answer 1

$$\begin{gathered} \mathrm{For}{x}^2-Px+8=0 \\ \mathrm{Let}\mathrm{us}\mathrm{suppose}\mathrm{two}\mathrm{roots}\mathrm{are}{z}_{1}and{z}_2 \\ \mathrm{given}:z_1-z_2=2 \\ \mathrm{also}{z}_1{z}_2=8\left(\mathrm{product}\mathrm{of}\mathrm{roots}\right) \\ \mathrm{i}.\mathrm{e}.\left(z_2+2\right)Z_2=8 \\ \mathrm{i}.\mathrm{e}.z_2^2+2z_2=8 \\ \mathrm{i}.\mathrm{e}.z_2^2+2z_2-8=0 \\ \mathrm{i}.\mathrm{e}.z_2^2+4z_2-2z_2-8=0 \\ \mathrm{i}.\mathrm{e}.\left(z_2+4\right)\left(z_2-2\right)=0 \\ \mathrm{i}.\mathrm{e}.z_2=-4\mathrm{or}{z}_2=2 \\ \Rightarrow z_1=-2\mathrm{or}{z}_2=4 \\ \mathrm{Since}P=z_1+z_2 \\ \Rightarrow P=-4-2\mathrm{or}P=4+2 \\ \mathrm{i}.\mathrm{e}.P=-6\mathrm{or}6 \end{gathered}$$