Question

Solve the following quadratics $21x^2+9x+1=0$

Answer 1

$21x^2+9x+1=0\text{Comparing the given, equation with the general form}a{x}^2+bx+c=0,wegeta=21,b=9,c=1Substitutingaandbin,\alpha =\frac{-b+\sqrt{b^{2-4ac}}}{2a}and\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}\alpha =\frac{-9+\sqrt{81-84}}{42}and\beta =\frac{-9-\sqrt{81-84}}{42}\Rightarrow \alpha =\frac{-9+\sqrt{-3}}{42}and\beta =\frac{-9-\sqrt{-3}}{42}\Rightarrow \alpha =\frac{-9+i\sqrt{3}}{42}and\beta =\frac{-9-i\sqrt{3}}{42}\text{The roots are}x=\frac{-9}{42}\pm \frac{i\sqrt{3}}{42}$