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

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

Solve for x (complex solution)

$$x=2$$

$$x=0$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Combine 2x and x to get 3x.

$$3x-2-2\sqrt{2x-2}\sqrt{x-2}-2=\left(\sqrt{x}\right)^{2}$$

Subtract 2 from -2 to get -4.

$$3x-4-2\sqrt{2x-2}\sqrt{x-2}=\left(\sqrt{x}\right)^{2}$$

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

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

Subtract 3x-4 from both sides of the equation.

$$-2\sqrt{2x-2}\sqrt{x-2}=x-\left(3x-4\right)$$

To find the opposite of 3x-4, find the opposite of each term.

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

Combine x and -3x to get -2x.

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

Square both sides of the equation.

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

Expand \left(-2\sqrt{2x-2}\sqrt{x-2}\right)^{2}.

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

Calculate -2 to the power of 2 and get 4.

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

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

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

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

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

Use the distributive property to multiply 4 by 2x-2.

$$\left(8x-8\right)\left(x-2\right)=\left(-2x+4\right)^{2}$$

Apply the distributive property by multiplying each term of 8x-8 by each term of x-2.

$$8x^{2}-16x-8x+16=\left(-2x+4\right)^{2}$$

Combine -16x and -8x to get -24x.

$$8x^{2}-24x+16=\left(-2x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2x+4\right)^{2}.

$$8x^{2}-24x+16=4x^{2}-16x+16$$

Subtract 4x^{2} from both sides.

$$8x^{2}-24x+16-4x^{2}=-16x+16$$

Combine 8x^{2} and -4x^{2} to get 4x^{2}.

$$4x^{2}-24x+16=-16x+16$$

Add 16x to both sides.

$$4x^{2}-24x+16+16x=16$$

Combine -24x and 16x to get -8x.

$$4x^{2}-8x+16=16$$

Subtract 16 from both sides.

$$4x^{2}-8x+16-16=0$$

Subtract 16 from 16 to get 0.

$$4x^{2}-8x=0$$

Factor out x.

$$x\left(4x-8\right)=0$$

To find equation solutions, solve x=0 and 4x-8=0.

$$x=0 x=2$$

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

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

Simplify. The value x=0 satisfies the equation.

$$0=0$$

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

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

Simplify. The value x=2 satisfies the equation.

$$2^{\frac{1}{2}}=2^{\frac{1}{2}}$$

List all solutions of \sqrt{2x-2}-\sqrt{x-2}=\sqrt{x}.

$$x=0 x=2$$

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Combine 2x and x to get 3x.

$$3x-2-2\sqrt{2x-2}\sqrt{x-2}-2=\left(\sqrt{x}\right)^{2}$$

Subtract 2 from -2 to get -4.

$$3x-4-2\sqrt{2x-2}\sqrt{x-2}=\left(\sqrt{x}\right)^{2}$$

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

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

Subtract 3x-4 from both sides of the equation.

$$-2\sqrt{2x-2}\sqrt{x-2}=x-\left(3x-4\right)$$

To find the opposite of 3x-4, find the opposite of each term.

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

Combine x and -3x to get -2x.

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

Square both sides of the equation.

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

Expand \left(-2\sqrt{2x-2}\sqrt{x-2}\right)^{2}.

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

Calculate -2 to the power of 2 and get 4.

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

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

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

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

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

Use the distributive property to multiply 4 by 2x-2.

$$\left(8x-8\right)\left(x-2\right)=\left(-2x+4\right)^{2}$$

Apply the distributive property by multiplying each term of 8x-8 by each term of x-2.

$$8x^{2}-16x-8x+16=\left(-2x+4\right)^{2}$$

Combine -16x and -8x to get -24x.

$$8x^{2}-24x+16=\left(-2x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2x+4\right)^{2}.

$$8x^{2}-24x+16=4x^{2}-16x+16$$

Subtract 4x^{2} from both sides.

$$8x^{2}-24x+16-4x^{2}=-16x+16$$

Combine 8x^{2} and -4x^{2} to get 4x^{2}.

$$4x^{2}-24x+16=-16x+16$$

Add 16x to both sides.

$$4x^{2}-24x+16+16x=16$$

Combine -24x and 16x to get -8x.

$$4x^{2}-8x+16=16$$

Subtract 16 from both sides.

$$4x^{2}-8x+16-16=0$$

Subtract 16 from 16 to get 0.

$$4x^{2}-8x=0$$

Factor out x.

$$x\left(4x-8\right)=0$$

To find equation solutions, solve x=0 and 4x-8=0.

$$x=0 x=2$$

Substitute 0 for x in the equation \sqrt{2x-2}-\sqrt{x-2}=\sqrt{x}. The expression \sqrt{2\times 0-2} is undefined because the radicand cannot be negative.

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

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

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

Simplify. The value x=2 satisfies the equation.

$$2^{\frac{1}{2}}=2^{\frac{1}{2}}$$

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

$$x=2$$