Solution
Given
[tex]\begin{gathered} x^4-20x^2=-64 \\ \\ Factor\text{ x}^2\text{ out} \\ x^2(x^2-20)=\text{ -64} \\ Let\text{ x}^2\text{ be represented by p}^ \\ x^2(x^2-20)=\text{-64 becomes} \\ p(p-20)=-64 \\ p^2-20p=-64 \\ p^2-20p+64=0 \end{gathered}[/tex][tex]\begin{gathered} p^2-16p-4p+64=0 \\ p(p-16)-4(p-16)=0 \\ (p-16)(p-4)=0 \\ p-16=0\text{ and p-4=0} \\ p=16\text{ and p=4} \end{gathered}[/tex][tex]\begin{gathered} Recall\text{ that x}^2=p \\ when\text{ p = 16} \\ x^2=16 \\ x=\pm\sqrt{16} \\ x=\pm4 \end{gathered}[/tex][tex]\begin{gathered} When\text{ p}=4 \\ x^2=4 \\ x=\pm\sqrt{4} \\ x=\pm2 \end{gathered}[/tex]
Thus, the real solutions of the equation are x = 4, x= -4, x = 2, x = -2
