Question

$x^2-6x+4=0$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ {x}^2-6x+4=0 \\ \mathrm{On}\mathrm{c}\text{omparing it with}a{x}^2+bx+c=0,\text{we get:} \\ a=1,b=-6\text{and}c=4 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ \text{=}{(-6)}^2-4\times 1\times 4 \\ \text{= 36}-16 \\ \text{= 20 > 0} \\ \text{Hence, the roots of the equation are real.} \\ \mathrm{Roots}\alpha \mathrm{and}\beta \mathrm{are}\mathrm{given}\mathrm{by}: \\ \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-(-6)+\sqrt{20}}{2\times 1}=\frac{6+2\sqrt{5}}{2}=\frac{2(3+\sqrt{5})}{2}\text{=}(3+\sqrt{5}) \\ \begin{array}{l}\beta =\frac{-b-\sqrt{D}}{2a}=\frac{-(-6)-\sqrt{20}}{2\times 1}=\frac{6-2\sqrt{5}}{2}=\frac{2(3-\sqrt{5})}{2}=(3-\sqrt{5})\\ \text{Thus, the roots of the equation are}(3+2\sqrt{5})\text{and}(3-2\sqrt{5})\text{.}\end{array} \end{gathered}$$