An inefficient root finder

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

$x=\left( 1+i \right) \sqrt [ 4 ]{ \frac { 3 }{ 2 } { x }^{ 2 }+C }$

Raising both sides to fourth power (Also note $(1+i)^4=((1+i)^2)^2=(2i)^2=-4$ ):-

$x^4=(-4)\left(\dfrac{3x^2}{2}+C\right)$

$\implies x^4+6x^2+4C=0$

Comparing it with $(x^2+A)(x^2+B)=0$ , we get $A+B=6$ , solving it with $B-A=456$ by adding and subtracting them respectively we get $B=231,A=-225$ . Also by comparison we get $4C=AB=-(225)(231)$ . Hence the equation is:

$x^2+6x-(225)(231)=0$

$\implies (x^2+3)^2=9+\underbrace{(225)(231)}_{(228-3)(228+3)}$

$\implies (x^2+3)^2=\not9+(228^2-\not9)=228^2$

$\implies x^2+3=228$

$\implies x^2=225\implies \boxed{x=15}(**)$

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( $(**)x=-15$ doesn't satisfy $x=\left( 1+i \right) \sqrt [ 4 ]{ \frac { 3 }{ 2 } { x }^{ 2 }+\underbrace{C }_{\frac{-(225)(231)}4}}~~$ )

** $I$ $didn't$ $get$ $what$ $made$ $this$ $problem$ $rise$ $to$ $lvl 5???**$