Question

Solve using “completing the square”, thank you! x^4-4x^2+2=0

Answer 1

$(*)\qquad(a-b)^2=a^2-2ab+b^2\\\\x^4=x^{2\cdot2}=(x^2)^2\\\\x^4-4x^2+2=0\\\\(x^2)^2-4x^2+2=0\\\\\text{substitute}\ t=x^2 > 0\\\\t^2-4t+2=0\\\\t^2-2\cdot t\cdot2+2=0\ \ \ \ \ \ |+2^2\\\\\underbrace{t^2-2\cdot t\cdot2+2^2}_{(*)}+2=2^2\ \ \ \ \ \ |-2\\\\(t-2)^2=4-2\\\\(t-2)^2=2\to t-2=\pm\sqrt2\ \ \ \ \ |+2\\\\t=2-\sqrt2\ \vee\ t=2+\sqrt2\\\\x^2=2-\sqrt2\ \vee\ x^2=2+\sqrt2\\\\x=-\sqrt{2-\sqrt2}\ \vee\ x=\sqrt{2-\sqrt2}\ \vee\ x=-\sqrt{2+\sqrt2}\ \vee\ x=\sqrt{2+\sqrt2}$