Question

Find the value of P for which the quadratic equation 2px2-8px+1=0 has equal roots?

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{is}, \\ 2{\mathrm{px}}^2-8\mathrm{px}+1=0 \\ \mathrm{Now},\mathrm{a}=2\mathrm{p};\mathrm{b}=-8\mathrm{p};\mathrm{c}=1 \\ \mathrm{Now},\mathrm{D}={\mathrm{b}}^2-4\mathrm{ac}={\left(-8\mathrm{p}\right)}^2-4\left(2\mathrm{p}\right)\left(1\right)=64{\mathrm{p}}^2-8\mathrm{p} \\ \mathrm{Now},\mathrm{for}\mathrm{equal}\mathrm{roots}, \\ \mathrm{D}=0 \\ \Rightarrow 64{\mathrm{p}}^2-8\mathrm{p}=0 \\ \Rightarrow 8\mathrm{p}\left(8\mathrm{p}-1\right)=0 \\ \Rightarrow \mathrm{p}=0\mathrm{or}8\mathrm{p}-1=0 \\ \Rightarrow \mathrm{p}=0\left(\mathrm{rejected}\right)\mathrm{or}\mathrm{p}=\frac{1}{8} \\ \mathrm{Hence},\mathrm{p}=\frac{1}{8}. \end{gathered}$$