Solve sqrt{6+x}+sqrt{x+2}=2sqrt{3+x}

$$\sqrt { 6 + x } + \sqrt { x + 2 } = 2 \sqrt { 3 + x }$$

Solve for x

$$x=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{6+x}+\sqrt{x+2}\right)^{2}=\left(2\sqrt{3+x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6+x}+\sqrt{x+2}\right)^{2}.

$$\left(\sqrt{6+x}\right)^{2}+2\sqrt{6+x}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}=\left(2\sqrt{3+x}\right)^{2}$$

Calculate \sqrt{6+x} to the power of 2 and get 6+x.

$$6+x+2\sqrt{6+x}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}=\left(2\sqrt{3+x}\right)^{2}$$

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

$$6+x+2\sqrt{6+x}\sqrt{x+2}+x+2=\left(2\sqrt{3+x}\right)^{2}$$

Combine x and x to get 2x.

$$6+2x+2\sqrt{6+x}\sqrt{x+2}+2=\left(2\sqrt{3+x}\right)^{2}$$

Add 6 and 2 to get 8.

$$8+2x+2\sqrt{6+x}\sqrt{x+2}=\left(2\sqrt{3+x}\right)^{2}$$

Expand \left(2\sqrt{3+x}\right)^{2}.

$$8+2x+2\sqrt{6+x}\sqrt{x+2}=2^{2}\left(\sqrt{3+x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$8+2x+2\sqrt{6+x}\sqrt{x+2}=4\left(\sqrt{3+x}\right)^{2}$$

Calculate \sqrt{3+x} to the power of 2 and get 3+x.

$$8+2x+2\sqrt{6+x}\sqrt{x+2}=4\left(3+x\right)$$

Use the distributive property to multiply 4 by 3+x.

$$8+2x+2\sqrt{6+x}\sqrt{x+2}=12+4x$$

Subtract 8+2x from both sides of the equation.

$$2\sqrt{6+x}\sqrt{x+2}=12+4x-\left(8+2x\right)$$

To find the opposite of 8+2x, find the opposite of each term.

$$2\sqrt{6+x}\sqrt{x+2}=12+4x-8-2x$$

Subtract 8 from 12 to get 4.

$$2\sqrt{6+x}\sqrt{x+2}=4+4x-2x$$

Combine 4x and -2x to get 2x.

$$2\sqrt{6+x}\sqrt{x+2}=4+2x$$

Square both sides of the equation.

$$\left(2\sqrt{6+x}\sqrt{x+2}\right)^{2}=\left(4+2x\right)^{2}$$

Expand \left(2\sqrt{6+x}\sqrt{x+2}\right)^{2}.

$$2^{2}\left(\sqrt{6+x}\right)^{2}\left(\sqrt{x+2}\right)^{2}=\left(4+2x\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{6+x}\right)^{2}\left(\sqrt{x+2}\right)^{2}=\left(4+2x\right)^{2}$$

Calculate \sqrt{6+x} to the power of 2 and get 6+x.

$$4\left(6+x\right)\left(\sqrt{x+2}\right)^{2}=\left(4+2x\right)^{2}$$

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

$$4\left(6+x\right)\left(x+2\right)=\left(4+2x\right)^{2}$$

Use the distributive property to multiply 4 by 6+x.

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

Apply the distributive property by multiplying each term of 24+4x by each term of x+2.

$$24x+48+4x^{2}+8x=\left(4+2x\right)^{2}$$

Combine 24x and 8x to get 32x.

$$32x+48+4x^{2}=\left(4+2x\right)^{2}$$

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

$$32x+48+4x^{2}=16+16x+4x^{2}$$

Subtract 16x from both sides.

$$32x+48+4x^{2}-16x=16+4x^{2}$$

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

$$16x+48+4x^{2}=16+4x^{2}$$

Subtract 4x^{2} from both sides.

$$16x+48+4x^{2}-4x^{2}=16$$

Combine 4x^{2} and -4x^{2} to get 0.

$$16x+48=16$$

Subtract 48 from both sides.

$$16x=16-48$$

Subtract 48 from 16 to get -32.

$$16x=-32$$

Divide both sides by 16.

$$x=\frac{-32}{16}$$

Divide -32 by 16 to get -2.

$$x=-2$$

Substitute -2 for x in the equation \sqrt{6+x}+\sqrt{x+2}=2\sqrt{3+x}.

$$\sqrt{6-2}+\sqrt{-2+2}=2\sqrt{3-2}$$

Simplify. The value x=-2 satisfies the equation.

$$2=2$$

Equation \sqrt{x+2}+\sqrt{x+6}=2\sqrt{x+3} has a unique solution.

$$x=-2$$