Question

Show that the quadratic equation x² – 8x + 18 = 0 has no real solution.

Answer 1

Given: x² – 8x + 18 = 0

$$\begin{gathered} x^2-8x+18=0 \\ \mathrm{Adding}\mathrm{and}\mathrm{subtracting}{\left(\frac{1}{2}\times 8\right)}^2,\mathrm{we}\mathrm{get} \\ \Rightarrow x^2-8x+18+4^2-4^2=0 \\ \Rightarrow x^2-8x+16+18-16=0 \\ \Rightarrow {\left(x-4\right)}^2+2=0 \\ \Rightarrow {\left(x-4\right)}^2=-2 \\ \Rightarrow \left(x-4\right)=\pm \sqrt{-2} \\ \mathrm{But},\sqrt{-2}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{real}\mathrm{number}. \end{gathered}$$

Hence, the quadratic equation x² – 8x + 18 = 0 has no real solution.