Solve b=x^4-2(x+1)^2+4x+3

$$b = x ^ { 4 } - 2 ( x + 1 ) ^ { 2 } + 4 x + 3$$

Solve for b

$$b=\left(x^{2}-1\right)^{2}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Use the distributive property to multiply -2 by x^{2}+2x+1.

$$b=x^{4}-2x^{2}-4x-2+4x+3$$

Combine -4x and 4x to get 0.

$$b=x^{4}-2x^{2}-2+3$$

Add -2 and 3 to get 1.

$$b=x^{4}-2x^{2}+1$$

Solve for x (complex solution)

$$x=\sqrt{-\sqrt{b}+1}$$

$$x=-\sqrt{-\sqrt{b}+1}$$

$$x=-\sqrt{\sqrt{b}+1}$$

$$x=\sqrt{\sqrt{b}+1}$$

Solve for x

$$\left\{\begin{matrix}x=-\sqrt{\sqrt{b}+1}\text{; }x=\sqrt{\sqrt{b}+1}\text{, }&b\geq 0\\x=-\sqrt{-\sqrt{b}+1}\text{; }x=\sqrt{-\sqrt{b}+1}\text{, }&b\geq 0\text{ and }b\leq 1\end{matrix}\right.$$