Solve sqrt{x}+sqrt{x+4}=sqrt{2x+4}

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

Solve for x (complex solution)

$$x=-4$$

$$x=0$$

Solution Steps

Subtract \sqrt{x+4} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

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

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

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

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

Combine 2x and x to get 3x.

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

Add 4 and 4 to get 8.

$$x=3x+8-2\sqrt{2x+4}\sqrt{x+4}$$

Subtract 3x+8 from both sides of the equation.

$$x-\left(3x+8\right)=-2\sqrt{2x+4}\sqrt{x+4}$$

To find the opposite of 3x+8, find the opposite of each term.

$$x-3x-8=-2\sqrt{2x+4}\sqrt{x+4}$$

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

$$-2x-8=-2\sqrt{2x+4}\sqrt{x+4}$$

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2x-8\right)^{2}.

$$4x^{2}+32x+64=\left(-2\sqrt{2x+4}\sqrt{x+4}\right)^{2}$$

Expand \left(-2\sqrt{2x+4}\sqrt{x+4}\right)^{2}.

$$4x^{2}+32x+64=\left(-2\right)^{2}\left(\sqrt{2x+4}\right)^{2}\left(\sqrt{x+4}\right)^{2}$$

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

$$4x^{2}+32x+64=4\left(\sqrt{2x+4}\right)^{2}\left(\sqrt{x+4}\right)^{2}$$

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

$$4x^{2}+32x+64=4\left(2x+4\right)\left(\sqrt{x+4}\right)^{2}$$

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

$$4x^{2}+32x+64=4\left(2x+4\right)\left(x+4\right)$$

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

$$4x^{2}+32x+64=\left(8x+16\right)\left(x+4\right)$$

Apply the distributive property by multiplying each term of 8x+16 by each term of x+4.

$$4x^{2}+32x+64=8x^{2}+32x+16x+64$$

Combine 32x and 16x to get 48x.

$$4x^{2}+32x+64=8x^{2}+48x+64$$

Subtract 8x^{2} from both sides.

$$4x^{2}+32x+64-8x^{2}=48x+64$$

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

$$-4x^{2}+32x+64=48x+64$$

Subtract 48x from both sides.

$$-4x^{2}+32x+64-48x=64$$

Combine 32x and -48x to get -16x.

$$-4x^{2}-16x+64=64$$

Subtract 64 from both sides.

$$-4x^{2}-16x+64-64=0$$

Subtract 64 from 64 to get 0.

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

Factor out x.

$$x\left(-4x-16\right)=0$$

To find equation solutions, solve x=0 and -4x-16=0.

$$x=0 x=-4$$

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

$$\sqrt{0}+\sqrt{0+4}=\sqrt{2\times 0+4}$$

Simplify. The value x=0 satisfies the equation.

$$2=2$$

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

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

Simplify. The value x=-4 satisfies the equation.

$$2i=2i$$

List all solutions of \sqrt{x}=\sqrt{2x+4}-\sqrt{x+4}.

$$x=0 x=-4$$

Solve for x

$$x=0$$

Solution Steps

Subtract \sqrt{x+4} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

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

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

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

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

Combine 2x and x to get 3x.

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

Add 4 and 4 to get 8.

$$x=3x+8-2\sqrt{2x+4}\sqrt{x+4}$$

Subtract 3x+8 from both sides of the equation.

$$x-\left(3x+8\right)=-2\sqrt{2x+4}\sqrt{x+4}$$

To find the opposite of 3x+8, find the opposite of each term.

$$x-3x-8=-2\sqrt{2x+4}\sqrt{x+4}$$

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

$$-2x-8=-2\sqrt{2x+4}\sqrt{x+4}$$

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2x-8\right)^{2}.

$$4x^{2}+32x+64=\left(-2\sqrt{2x+4}\sqrt{x+4}\right)^{2}$$

Expand \left(-2\sqrt{2x+4}\sqrt{x+4}\right)^{2}.

$$4x^{2}+32x+64=\left(-2\right)^{2}\left(\sqrt{2x+4}\right)^{2}\left(\sqrt{x+4}\right)^{2}$$

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

$$4x^{2}+32x+64=4\left(\sqrt{2x+4}\right)^{2}\left(\sqrt{x+4}\right)^{2}$$

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

$$4x^{2}+32x+64=4\left(2x+4\right)\left(\sqrt{x+4}\right)^{2}$$

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

$$4x^{2}+32x+64=4\left(2x+4\right)\left(x+4\right)$$

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

$$4x^{2}+32x+64=\left(8x+16\right)\left(x+4\right)$$

Apply the distributive property by multiplying each term of 8x+16 by each term of x+4.

$$4x^{2}+32x+64=8x^{2}+32x+16x+64$$

Combine 32x and 16x to get 48x.

$$4x^{2}+32x+64=8x^{2}+48x+64$$

Subtract 8x^{2} from both sides.

$$4x^{2}+32x+64-8x^{2}=48x+64$$

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

$$-4x^{2}+32x+64=48x+64$$

Subtract 48x from both sides.

$$-4x^{2}+32x+64-48x=64$$

Combine 32x and -48x to get -16x.

$$-4x^{2}-16x+64=64$$

Subtract 64 from both sides.

$$-4x^{2}-16x+64-64=0$$

Subtract 64 from 64 to get 0.

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

Factor out x.

$$x\left(-4x-16\right)=0$$

To find equation solutions, solve x=0 and -4x-16=0.

$$x=0 x=-4$$

Substitute -4 for x in the equation \sqrt{x}+\sqrt{x+4}=\sqrt{2x+4}. The expression \sqrt{-4} is undefined because the radicand cannot be negative.

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

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

$$\sqrt{0}+\sqrt{0+4}=\sqrt{2\times 0+4}$$

Simplify. The value x=0 satisfies the equation.

$$2=2$$

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

$$x=0$$