Solve (2+sqrt{3})^2+(2-sqrt{3})^2-4=0

$$( 2 + \sqrt { 3 } ) ^ { 2 } + ( 2 - \sqrt { 3 } ) ^ { 2 } - 4 = 0$$

Verify

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{3}\right)^{2}.

$$4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}+\left(2-\sqrt{3}\right)^{2}-4=0$$

The square of \sqrt{3} is 3.

$$4+4\sqrt{3}+3+\left(2-\sqrt{3}\right)^{2}-4=0$$

Add 4 and 3 to get 7.

$$7+4\sqrt{3}+\left(2-\sqrt{3}\right)^{2}-4=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-\sqrt{3}\right)^{2}.

$$7+4\sqrt{3}+4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}-4=0$$

The square of \sqrt{3} is 3.

$$7+4\sqrt{3}+4-4\sqrt{3}+3-4=0$$

Add 4 and 3 to get 7.

$$7+4\sqrt{3}+7-4\sqrt{3}-4=0$$

Add 7 and 7 to get 14.

$$14+4\sqrt{3}-4\sqrt{3}-4=0$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$14-4=0$$

Subtract 4 from 14 to get 10.

$$10=0$$

Compare 10 and 0.

$$\text{false}$$