Question

x^4-4x^2-1=0
Solve for x

Answer 1

Factorize this, this is a perfect square a^2-2ab+b^2=(a-b)^2

Answer 2

Put \( \large u=x^2\), then
\[
\large u^2 - 4 u -1=0

\]
Use the quadratic formula to find u, then find x

Answer 3

\[
u=\frac{4+ \sqrt{16+4}}{2}=\frac{1}{2}
\left(2 \sqrt{5}+4\right)=2+\sqrt 5>0\\
u=\frac{4- \sqrt{16+4}}{2\ }=\frac{1}{2}
\left(2 \sqrt{5}-4\right)=-2+ \sqrt 5<0\\
\]

Answer 4

Since u is positive, we take the first solution. Solve for x to get
\[
\left\{x=
-\sqrt{2+\sqrt{5}}\right\},\left\{x=
\sqrt{2+\sqrt{5}}\right\}
\]