It is given that y1 exceeds y2 by 28, it implies that:
[tex]y_1=y_2+28[/tex]
Substitute the expressions for y1 and y2 into the equation:
[tex]\begin{gathered} (x^2-2)^2=12(x^2-2) \\ \end{gathered}[/tex]
Expand the expressions on both sides and solve the equation for x:
[tex]\begin{gathered} x^4-4x^2+4=12x^2-24 \\ \Rightarrow x^4-4x^2-12x^2_{}+4+24=0 \\ \Rightarrow x^4-16x^2+28=0 \\ \text{let }x^2=y\text{ in the equation:} \\ \Rightarrow y^2-16y+28=0 \\ \Rightarrow y^2-14y-2y+28=0 \\ \Rightarrow y(y-14)-2(y-14)=0 \\ \Rightarrow(y-14)(y-2)=0 \\ \Rightarrow y=14,2 \end{gathered}[/tex]
Substitute the values of y back into the equation x²=y:
[tex]\begin{gathered} x^2=14\text{ and }x^2=2 \\ \Rightarrow x=\pm\sqrt[]{14},\pm\sqrt[]{2} \\ \text{The solution set is }\times\mleft\lbrace-\sqrt[]{14},-\sqrt[]{2,}\sqrt[]{2},\sqrt[]{14}\mright\rbrace \end{gathered}[/tex]
