Solve 2x+1+x^2=left(x+1right)^2

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

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(x+1\right)^{2}.

$$2x+1+x^{2}=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$2x+1+x^{2}-x^{2}=2x+1$$

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

$$2x+1=2x+1$$

Subtract 2x from both sides.

$$2x+1-2x=1$$

Combine 2x and -2x to get 0.

$$1=1$$

Compare 1 and 1.

$$\text{true}$$

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(x+1\right)^{2}.

$$2x+1+x^{2}=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$2x+1+x^{2}-x^{2}=2x+1$$

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

$$2x+1=2x+1$$

Subtract 2x from both sides.

$$2x+1-2x=1$$

Combine 2x and -2x to get 0.

$$1=1$$

Compare 1 and 1.

$$\text{true}$$

This is true for any x.

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