Question

Solve the equation $x^2+x+\frac{1}{\sqrt{2}}=0$

Answer 1

Step 1 : Given $x^2+x+\frac{1}{\sqrt{2}}=0\Rightarrow \sqrt{2}{x}^2+\sqrt{2}x+1=0$

Step 2: $D=b^2-4aca=\sqrt{2},b=\sqrt{2},c=1$
$\Rightarrow D=(\sqrt{2}{)}^2-4(\sqrt{2}\left)\right(1)=2-4\sqrt{2}<0\therefore D$ is negative
$\therefore$ Roots are imaginary, so we can use the quadratic formula.

Step 3: Using quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-\sqrt{2}\pm (\sqrt{(\sqrt{2}{)}^2-4(\sqrt{2}\left)\right(1)}}{2\sqrt{2}}$

$\Rightarrow x=\frac{-\sqrt{2}\pm \sqrt{2(1-2\sqrt{2}})}{2\sqrt{2}}$

$\Rightarrow x=\frac{-\sqrt{2}\pm \sqrt{2-4\sqrt{2}}}{2\sqrt{2}}$

$\Rightarrow x=\frac{-1\pm \sqrt{-(2\sqrt{2}-1)}}{2}\Rightarrow x=\frac{-1\pm \sqrt{(2\sqrt{2}-1)}i}{2}$

The values of x are x=$\frac{-1+\sqrt{(2\sqrt{2}-1)}i}{2}$ and $\frac{-1-\sqrt{(2\sqrt{2}-1)}i}{2}$
Root of the quadratic equation.