Question

If α, β are the zeros of the polynomial f(x) = x² − 5x + k such that α − β = 1, find the value of k.

Answer 1

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=x^2-5x+k \\ \mathrm{The}\mathrm{co}-\mathrm{efficients}\mathrm{are}a=1,b=-5\mathrm{and}c=k. \\ \therefore \alpha +\beta =\frac{-b}{a} \\ =>\alpha +\beta =-\frac{(-5)}{1} \\ =>\alpha +\beta =5\dots \left(1\right) \\ \mathrm{Also},\alpha -\beta =1\dots \left(2\right) \\ \mathrm{From}\left(1\right)\&\left(2\right),\mathrm{we}\mathrm{}\mathrm{get}: \\ 2\mathrm{\alpha }=6 \\ =>\mathrm{\alpha }=3 \\ \mathrm{Pu}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{\alpha }\mathrm{in}\left(1\right),\mathrm{w}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{e}\mathrm{t}\mathrm{}\mathrm{\beta }=2. \\ \mathrm{Now},\mathrm{\alpha }\mathrm{\beta }=\frac{c}{a} \\ =>3\times 2=\frac{k}{1} \\ \therefore k=6 \end{gathered}$$