Question

Determine the values of p for which the quadratic equation $2x^2+px+8=0$ has real roots.

Answer 1

$$\begin{gathered} \text{Given:} \\ 2x^2+px+8=0 \\ \text{Here,} \\ a=2,b=p\mathrm{and}c=8 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ \text{=}{p}^2-4\times 2\times 8 \\ \text{=}(p^2-64) \\ \mathrm{If}D\ge 0,\mathrm{t}\text{he roots of the equation will be real.} \\ \Rightarrow (p^2-64)\ge 0 \\ \Rightarrow (p+8)(p-8)\ge 0 \\ \Rightarrow p\ge 8\mathrm{and}p\le -8 \\ \mathrm{T}\text{hus, the roots of the equation are real for}p\ge 8\mathrm{and}p\le -8. \end{gathered}$$