Solve alpha^2+beta^2=(alpha+beta)^2-2alphabeta

$$\alpha ^ { 2 } + \beta ^ { 2 } = ( \alpha + \beta ) ^ { 2 } - 2 \alpha \beta$$

Solve for α

$$\alpha \in \mathrm{R}$$

Solution Steps

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

$$\alpha ^{2}+\beta ^{2}=\alpha ^{2}+2\alpha \beta +\beta ^{2}-2\alpha \beta$$

Combine 2\alpha \beta and -2\alpha \beta to get 0.

$$\alpha ^{2}+\beta ^{2}=\alpha ^{2}+\beta ^{2}$$

Subtract \alpha ^{2} from both sides.

$$\alpha ^{2}+\beta ^{2}-\alpha ^{2}=\beta ^{2}$$

Combine \alpha ^{2} and -\alpha ^{2} to get 0.

$$\beta ^{2}=\beta ^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any \alpha .

$$\alpha \in \mathrm{R}$$

Solve for β

$$\beta \in \mathrm{R}$$

Solution Steps

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

$$\alpha ^{2}+\beta ^{2}=\alpha ^{2}+2\alpha \beta +\beta ^{2}-2\alpha \beta$$

Combine 2\alpha \beta and -2\alpha \beta to get 0.

$$\alpha ^{2}+\beta ^{2}=\alpha ^{2}+\beta ^{2}$$

Subtract \beta ^{2} from both sides.

$$\alpha ^{2}+\beta ^{2}-\beta ^{2}=\alpha ^{2}$$

Combine \beta ^{2} and -\beta ^{2} to get 0.

$$\alpha ^{2}=\alpha ^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any \beta .

$$\beta \in \mathrm{R}$$