Solve sqrt{2-x}=x-2/2

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

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

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

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

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

To raise \frac{x-2}{2} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Divide each term of x^{2}-4x+4 by 4 to get \frac{1}{4}x^{2}-x+1.

$$2-x=\frac{1}{4}x^{2}-x+1$$

Subtract \frac{1}{4}x^{2} from both sides.

$$2-x-\frac{1}{4}x^{2}=-x+1$$

Add x to both sides.

$$2-x-\frac{1}{4}x^{2}+x=1$$

Combine -x and x to get 0.

$$2-\frac{1}{4}x^{2}=1$$

Subtract 2 from both sides.

$$-\frac{1}{4}x^{2}=1-2$$

Subtract 2 from 1 to get -1.

$$-\frac{1}{4}x^{2}=-1$$

Multiply both sides by -4, the reciprocal of -\frac{1}{4}.

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

Multiply -1 and -4 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

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

$$\sqrt{2-2}=\frac{2-2}{2}$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

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

$$\sqrt{2-\left(-2\right)}=\frac{-2-2}{2}$$

Simplify. The value x=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

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

$$x=2$$