Question

Find the number of real solutions of equation $x^4+x^2+1=0$.

Answer 1

Given equation

$x^4+x^2+1=0$

$\Rightarrow {\left(x^2\right)}^2+x^2+1=0$

Comparing that

$a{x}^2+bx+c=0$ and we get,

Then,

$a=1,b=1,c=1$

Using quadratic formula,

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

But given $X=x^2$ then

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

$\Rightarrow x^2=\frac{-1\pm \sqrt{1^2-4\times 1\times 1}}{2\times 1}$

$\Rightarrow x^2=\frac{-1\pm \sqrt{1-4}}{2}$

$\Rightarrow x^2=\frac{-1\pm \sqrt{-3}}{2}$

$\Rightarrow x^2=\frac{-1\pm \sqrt{-1}\times \sqrt{3}}{2}$

$\therefore 0$ real solutions