Solve 1*(2+x)^2=2^2+2(2)(x)+x^2

$$1 \cdot ( 2 + x ) ^ { 2 } = 2 ^ { 2 } + 2 ( 2 ) ( x ) + x ^ { 2 }$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

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

$$1\left(4+4x+x^{2}\right)=2^{2}+2\times 2x+x^{2}$$

Use the distributive property to multiply 1 by 4+4x+x^{2}.

$$4+4x+x^{2}=2^{2}+2\times 2x+x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4+4x+x^{2}=4+2\times 2x+x^{2}$$

Multiply 2 and 2 to get 4.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Subtract 4x from both sides.

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

Combine 4x and -4x to get 0.

$$x^{2}=x^{2}$$

Subtract x^{2} from both sides.

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

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

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

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

$$1\left(4+4x+x^{2}\right)=2^{2}+2\times 2x+x^{2}$$

Use the distributive property to multiply 1 by 4+4x+x^{2}.

$$4+4x+x^{2}=2^{2}+2\times 2x+x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4+4x+x^{2}=4+2\times 2x+x^{2}$$

Multiply 2 and 2 to get 4.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Subtract 4x from both sides.

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

Combine 4x and -4x to get 0.

$$x^{2}=x^{2}$$

Subtract x^{2} from both sides.

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

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

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$