Solve sqrt{2}*sqrt{18}-sqrt{27div3}-2-{left(sqrt{3}+1right)}^2

$$\sqrt{ 2 } \times \sqrt{ 18 } - \sqrt{ 27 \div 3 } -2- { \left( \sqrt{ 3 } +1 \right) }^{ 2 }$$

Evaluate

$$-2\sqrt{3}-3\approx -6.464101615$$

Solution Steps

Factor 18=2\times 9. Rewrite the square root of the product \sqrt{2\times 9} as the product of square roots \sqrt{2}\sqrt{9}.

$$\sqrt{2}\sqrt{2}\sqrt{9}-\sqrt{\frac{27}{3}}-2-\left(\sqrt{3}+1\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$2\sqrt{9}-\sqrt{\frac{27}{3}}-2-\left(\sqrt{3}+1\right)^{2}$$

Calculate the square root of 9 and get 3.

$$2\times 3-\sqrt{\frac{27}{3}}-2-\left(\sqrt{3}+1\right)^{2}$$

Multiply 2 and 3 to get 6.

$$6-\sqrt{\frac{27}{3}}-2-\left(\sqrt{3}+1\right)^{2}$$

Divide 27 by 3 to get 9.

$$6-\sqrt{9}-2-\left(\sqrt{3}+1\right)^{2}$$

Calculate the square root of 9 and get 3.

$$6-3-2-\left(\sqrt{3}+1\right)^{2}$$

Subtract 3 from 6 to get 3.

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

Subtract 2 from 3 to get 1.

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

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

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

The square of \sqrt{3} is 3.

$$1-\left(3+2\sqrt{3}+1\right)$$

Add 3 and 1 to get 4.

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

To find the opposite of 4+2\sqrt{3}, find the opposite of each term.

$$1-4-2\sqrt{3}$$

Subtract 4 from 1 to get -3.

$$-3-2\sqrt{3}$$