Solve (7+sqrt{3})(2-sqrt{3})^2+(2^2-3)+sqrt{3}

$$( 7 + \sqrt { 3 } ) ( 2 - \sqrt { 3 } ) ^ { 2 } + ( 2 ^ { 2 } - 3 ) + \sqrt { 3 }$$

Evaluate

$$38-20\sqrt{3}\approx 3.358983849$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

Use the distributive property to multiply 7+\sqrt{3} by 7-4\sqrt{3} and combine like terms.

$$49-21\sqrt{3}-4\left(\sqrt{3}\right)^{2}+2^{2}-3+\sqrt{3}$$

The square of \sqrt{3} is 3.

$$49-21\sqrt{3}-4\times 3+2^{2}-3+\sqrt{3}$$

Multiply -4 and 3 to get -12.

$$49-21\sqrt{3}-12+2^{2}-3+\sqrt{3}$$

Subtract 12 from 49 to get 37.

$$37-21\sqrt{3}+2^{2}-3+\sqrt{3}$$

Calculate 2 to the power of 2 and get 4.

$$37-21\sqrt{3}+4-3+\sqrt{3}$$

Add 37 and 4 to get 41.

$$41-21\sqrt{3}-3+\sqrt{3}$$

Subtract 3 from 41 to get 38.

$$38-21\sqrt{3}+\sqrt{3}$$

Combine -21\sqrt{3} and \sqrt{3} to get -20\sqrt{3}.

$$38-20\sqrt{3}$$