Solve sqrt{a-1}+left|b^2+x^2right|=x

$$\sqrt{ a-1 } + \left| { b }^{ 2 } + { x }^{ 2 } \right| = x$$

Solve for a

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

$$-x^{2}+x-b^{2}\geq 0$$

Steps for Solving Radical Equations

Subtract b^{2}+x^{2} from both sides of the equation.

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

Subtracting b^{2}+x^{2} from itself leaves 0.

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

Subtract b^{2}+x^{2} from x.

$$\sqrt{a-1}=-x^{2}+x-b^{2}$$

Square both sides of the equation.

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

Add 1 to both sides of the equation.

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

Subtracting -1 from itself leaves 0.

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

Subtract -1 from \left(-b^{2}-x^{2}+x\right)^{2}.

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

Solve for b

$$\left\{\begin{matrix}b=-\sqrt{-x^{2}-x+\sqrt{a-1}}\text{; }b=\sqrt{-x^{2}-x+\sqrt{a-1}}\text{, }&a>1\text{ and }x>0\text{ and }a\leq x^{2}+1\text{ and }a\geq x^{2}\left(x+1\right)^{2}+1\\b=-\sqrt{-x^{2}+x-\sqrt{a-1}}\text{; }b=\sqrt{-x^{2}+x-\sqrt{a-1}}\text{, }&x>0\text{ and }x<1\text{ and }a\leq x^{2}\left(x-1\right)^{2}+1\text{ and }a\geq 1\text{ and }a\leq x^{2}+1\end{matrix}\right.$$