[tex]\begin{gathered} \text{Given} \\ 4x^4=32x^3-68x^2 \end{gathered}[/tex]
Equate the to zero, such that all the terms on the right side is transposed to the left side.
[tex]\begin{gathered} 4x^4=32x^3-68x^2 \\ 4x^4-32x^3+68x^2=0 \\ \\ \text{Factor out all the GCF of all the terms which is }4x^2 \\ 4x^4-32x^3+68x^2=0 \\ 4x^2(x^2-8x+17)=0 \end{gathered}[/tex]
Apply the quadratic formula to the quadratic term.
[tex]\begin{gathered} x^2-8x+17=0 \\ \text{where} \\ a=1,b=-8,c=17 \\ \\ x=\frac{ -b \pm\sqrt{b^2 - 4ac}}{ 2a } \\ x=\frac{ -(-8) \pm\sqrt{(-8)^2 - 4(1)(17)}}{ 2(1) } \\ x=\frac{ 8 \pm\sqrt{64 - 68}}{ 2 } \\ x=\frac{ 8 \pm\sqrt{-4}}{ 2 } \\ x=\frac{ 8 \pm2\, i}{ 2 } \\ x=\frac{ 8 }{ 2 }\pm\frac{2\, i}{ 2 } \\ \\ x=\frac{ 8 }{ 2 }+\frac{2\, i}{ 2 } \\ x=\frac{8}{2}+\frac{2i}{2} \\ x=4+i \\ \\ x=\frac{8}{2}-\frac{2i}{2} \\ x=\frac{8}{2}-\frac{2i}{2} \\ x=4-i \end{gathered}[/tex]
Now we have 2 solutions, solve for the other solution which is the solution for 4x².
[tex]\begin{gathered} \text{Equate to zero} \\ 4x^2=0 \\ \frac{4x^2}{4}=\frac{0}{4} \\ x^2=0 \\ \sqrt[]{x^2}=\sqrt[]{0} \\ x=0 \end{gathered}[/tex]
Summarizing the solution.
[tex]\begin{gathered} \text{The real solution for }x\text{ is }0, \\ \text{and the imaginary solution are }4\pm1i \end{gathered}[/tex]
