Solve 2-x=sqrt{4-x}

$$2 - x = \sqrt { 4 - x }$$

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

$$\left(2-x\right)^{2}=\left(\sqrt{4-x}\right)^{2}$$

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

$$4-4x+x^{2}=\left(\sqrt{4-x}\right)^{2}$$

Calculate \sqrt{4-x} to the power of 2 and get 4-x.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Add x to both sides.

$$-4x+x^{2}+x=0$$

Combine -4x and x to get -3x.

$$-3x+x^{2}=0$$

Factor out x.

$$x\left(-3+x\right)=0$$

To find equation solutions, solve x=0 and -3+x=0.

$$x=0 x=3$$

Substitute 0 for x in the equation 2-x=\sqrt{4-x}.

$$2-0=\sqrt{4-0}$$

Simplify. The value x=0 satisfies the equation.

$$2=2$$

Substitute 3 for x in the equation 2-x=\sqrt{4-x}.

$$2-3=\sqrt{4-3}$$

Simplify. The value x=3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Equation 2-x=\sqrt{4-x} has a unique solution.

$$x=0$$