Solve (1+sqrt{3})^2-(4-sqrt{5})^2-sqrt{12}

$$( 1 + \sqrt { 3 } ) ^ { 2 } - ( 4 - \sqrt { 5 } ) ^ { 2 } - \sqrt { 12 }$$

Evaluate

$$8\sqrt{5}-17\approx 0.88854382$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{3}\right)^{2}.

$$1+2\sqrt{3}+\left(\sqrt{3}\right)^{2}-\left(4-\sqrt{5}\right)^{2}-\sqrt{12}$$

The square of \sqrt{3} is 3.

$$1+2\sqrt{3}+3-\left(4-\sqrt{5}\right)^{2}-\sqrt{12}$$

Add 1 and 3 to get 4.

$$4+2\sqrt{3}-\left(4-\sqrt{5}\right)^{2}-\sqrt{12}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{5}\right)^{2}.

$$4+2\sqrt{3}-\left(16-8\sqrt{5}+\left(\sqrt{5}\right)^{2}\right)-\sqrt{12}$$

The square of \sqrt{5} is 5.

$$4+2\sqrt{3}-\left(16-8\sqrt{5}+5\right)-\sqrt{12}$$

Add 16 and 5 to get 21.

$$4+2\sqrt{3}-\left(21-8\sqrt{5}\right)-\sqrt{12}$$

To find the opposite of 21-8\sqrt{5}, find the opposite of each term.

$$4+2\sqrt{3}-21+8\sqrt{5}-\sqrt{12}$$

Subtract 21 from 4 to get -17.

$$-17+2\sqrt{3}+8\sqrt{5}-\sqrt{12}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$-17+2\sqrt{3}+8\sqrt{5}-2\sqrt{3}$$

Combine 2\sqrt{3} and -2\sqrt{3} to get 0.

$$-17+8\sqrt{5}$$