Question

$$2 { \left(x-3 \right) }^{ 2 } { \left(x+1 \right) }^{ 2 } +(x-5)(x-3)+4( { x }^{ 2 } -5x+1)=4 { x }^{ 2 } -12$$

Answer

x=1.9372123718262,4.132982635498

Solution


Expand.
\[2{x}^{4}+4{x}^{3}+2{x}^{2}-12{x}^{3}-24{x}^{2}-12x+18{x}^{2}+36x+18+{x}^{2}-3x-5x+15+4{x}^{2}-20x+4=4{x}^{2}-12\]
Simplify  \(2{x}^{4}+4{x}^{3}+2{x}^{2}-12{x}^{3}-24{x}^{2}-12x+18{x}^{2}+36x+18+{x}^{2}-3x-5x+15+4{x}^{2}-20x+4\)  to  \(2{x}^{4}-8{x}^{3}+{x}^{2}-4x+37\).
\[2{x}^{4}-8{x}^{3}+{x}^{2}-4x+37=4{x}^{2}-12\]
Move all terms to one side.
\[2{x}^{4}-8{x}^{3}+{x}^{2}-4x+37-4{x}^{2}+12=0\]
Simplify  \(2{x}^{4}-8{x}^{3}+{x}^{2}-4x+37-4{x}^{2}+12\)  to  \(2{x}^{4}-8{x}^{3}-3{x}^{2}-4x+49\).
\[2{x}^{4}-8{x}^{3}-3{x}^{2}-4x+49=0\]
No root was found algebraically. However, the following root(s) were found by numerical methods.
\[x=1.937212,4.132983\]