Solve -3+2sqrt{3}-{left(sqrt{6}-sqrt{2}right)}^2+frac{sqrt{75}-sqrt{48}}{sqrt{3}}

$$-3+2 \sqrt{ 3 } - { \left( \sqrt{ 6 } - \sqrt{ 2 } \right) }^{ 2 } + \frac{ \sqrt{ 75 } - \sqrt{ 48 } }{ \sqrt{ 3 } }$$

Evaluate

$$2\left(3\sqrt{3}-5\right)\approx 0.392304845$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-\sqrt{2}\right)^{2}.

$$-3+2\sqrt{3}-\left(\left(\sqrt{6}\right)^{2}-2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

The square of \sqrt{6} is 6.

$$-3+2\sqrt{3}-\left(6-2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$-3+2\sqrt{3}-\left(6-2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

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

$$-3+2\sqrt{3}-\left(6-2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}\right)+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

Multiply -2 and 2 to get -4.

$$-3+2\sqrt{3}-\left(6-4\sqrt{3}+\left(\sqrt{2}\right)^{2}\right)+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

The square of \sqrt{2} is 2.

$$-3+2\sqrt{3}-\left(6-4\sqrt{3}+2\right)+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

Add 6 and 2 to get 8.

$$-3+2\sqrt{3}-\left(8-4\sqrt{3}\right)+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

To find the opposite of 8-4\sqrt{3}, find the opposite of each term.

$$-3+2\sqrt{3}-8+4\sqrt{3}+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

Subtract 8 from -3 to get -11.

$$-11+2\sqrt{3}+4\sqrt{3}+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

Combine 2\sqrt{3} and 4\sqrt{3} to get 6\sqrt{3}.

$$-11+6\sqrt{3}+\frac{\sqrt{75}-\sqrt{48}}{\sqrt{3}}$$

Factor 75=5^{2}\times 3. Rewrite the square root of the product \sqrt{5^{2}\times 3} as the product of square roots \sqrt{5^{2}}\sqrt{3}. Take the square root of 5^{2}.

$$-11+6\sqrt{3}+\frac{5\sqrt{3}-\sqrt{48}}{\sqrt{3}}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$-11+6\sqrt{3}+\frac{5\sqrt{3}-4\sqrt{3}}{\sqrt{3}}$$

Combine 5\sqrt{3} and -4\sqrt{3} to get \sqrt{3}.

$$-11+6\sqrt{3}+\frac{\sqrt{3}}{\sqrt{3}}$$

Rewrite the division of square roots \frac{\sqrt{3}}{\sqrt{3}} as the square root of the division \sqrt{\frac{3}{3}} and perform the division.

$$-11+6\sqrt{3}+\sqrt{1}$$

Calculate the square root of 1 and get 1.

$$-11+6\sqrt{3}+1$$

Add -11 and 1 to get -10.

$$-10+6\sqrt{3}$$