Solve sqrt{a-b}+|b-1|=0

$$\sqrt { a - b } + | b - 1 | = 0$$

Solve for a

$$a=b^{2}-b+1$$

$$-\left(1-b\right)\geq 0\text{ and }b\leq 1$$

Steps for Solving Radical Equations

Subtract |b-1| from both sides of the equation.

$$\sqrt{a-b}+|b-1|-|b-1|=-|b-1|$$

Subtracting |b-1| from itself leaves 0.

$$\sqrt{a-b}=-|b-1|$$

Square both sides of the equation.

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

Subtract -b from both sides of the equation.

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

Subtracting -b from itself leaves 0.

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

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

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

Solve for b

$$b=1$$

$$a=1$$