Question

Solve the equation $x^2+3x+9=0$

Answer 1

Step $1:$ Given $x^2+3x+9=0$

Step $2:$ Check nature of roots
$D=b^2-4ac$
$a=1,b=3,c=9$
$\Rightarrow D=9-36=-27<0$
$\therefore D$ is negative
$\therefore$ Roots are imaginary, so we can use the quadratic formula.

Step 3: Using quadratic formula
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
By putting the values of $a,b,c$
$x=\frac{-3\pm \left(\sqrt{9-36}\right)}{2}$
$x=\frac{-3\pm \sqrt{-27}}{2}$
$x=\frac{-3\pm 3\sqrt{3}i}{2}(\because \sqrt{-1}=i)$

The values of $x$ are
$x=\frac{-3+3\sqrt{3}i}{2},\frac{-3-3\sqrt{3i}}{2}$