Answer
The solutions are $x=\pm\sqrt{-1-\dfrac{\sqrt{2}}{2}}$ and $x=\pm\sqrt{-1+\dfrac{\sqrt{2}}{2}}$
Work Step by Step
$2x^{4}+4x^{2}+1=0$
Let $x^{2}$ be equal to $w$:
$w=x^{2}$ so $w^{2}=x^{4}$
Substitute $x^{4}$ by $w^{2}$ and $x^{2}$ by $w$ in the original equation:
$2w^{2}+4w+1=0$
Use the quadratic formula to solve this equation. The formula is $w=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=2$, $b=4$ and $c=1$. Substitute the known values into the formula and simplify:
$w=\dfrac{-4\pm\sqrt{4^{2}-4(2)(1)}}{2(2)}=\dfrac{-4\pm\sqrt{16-8}}{4}=...$
$...=\dfrac{-4\pm\sqrt{8}}{4}=\dfrac{-4\pm2\sqrt{2}}{4}=-1\pm\dfrac{\sqrt{2}}{2}$
So $w=-1+\dfrac{\sqrt{2}}{2}$ and $w=-1-\dfrac{\sqrt{2}}{2}$
Substitute $w$ back to $x^{2}$ and solve for $x$:
$w=-1+\dfrac{\sqrt{2}}{2}$
$x^{2}=-1+\dfrac{\sqrt{2}}{2}$
$x=\pm\sqrt{-1+\dfrac{\sqrt{2}}{2}}$
$w=-1-\dfrac{\sqrt{2}}{2}$
$x^{2}=-1-\dfrac{\sqrt{2}}{2}$
$x=\pm\sqrt{-1-\dfrac{\sqrt{2}}{2}}$
The solutions are $x=\pm\sqrt{-1-\dfrac{\sqrt{2}}{2}}$ and $x=\pm\sqrt{-1+\dfrac{\sqrt{2}}{2}}$
