Question

Solve for the roots in the equation below. x^4+3x^2-4=0

Answer 1

Answer:

$\large\boxed{x=-1\ or\ x=1}$

Step-by-step explanation:

$x^4+3x^2-4=0\\\\x^{(2)(2)}+3x^2-4=0\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(x^2)^2+3x^2-4=0\qquad\text{substitute}\ x^2=t\geq0x=\\\\t^2+3t-4=0\\\\t^2+4t-t-4=0\\\\t(t+4)-1(t+4)=0\\\\(t+4)(t-1)=0\iff t+4=0\ \vee\ t-1=0\\\\t+4=0\qquad\text{subtract 4 from both sides}\\t=-40\to x^2=1\\\\x^2=1\to x=\pm\sqrt1\to x=-1\ \vee\ x=1$