Solve frac{sqrt{72}-sqrt{24}}{sqrt{8}}-(2+sqrt{3})^2

$$\frac { \sqrt { 72 } - \sqrt { 24 } } { \sqrt { 8 } } - ( 2 + \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$-5\sqrt{3}-4\approx -12.660254038$$

Short Solution Steps

Factor 72=6^{2}\times 2. Rewrite the square root of the product \sqrt{6^{2}\times 2} as the product of square roots \sqrt{6^{2}}\sqrt{2}. Take the square root of 6^{2}.

$$\frac{6\sqrt{2}-\sqrt{24}}{\sqrt{8}}-\left(2+\sqrt{3}\right)^{2}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$\frac{6\sqrt{2}-2\sqrt{6}}{\sqrt{8}}-\left(2+\sqrt{3}\right)^{2}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Rationalize the denominator of \frac{6\sqrt{2}-2\sqrt{6}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

$$\frac{\left(6\sqrt{2}-2\sqrt{6}\right)\sqrt{2}}{4}-\left(7+4\sqrt{3}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7+4\sqrt{3} times \frac{4}{4}.

$$\frac{\left(6\sqrt{2}-2\sqrt{6}\right)\sqrt{2}}{4}-\frac{4\left(7+4\sqrt{3}\right)}{4}$$

Since \frac{\left(6\sqrt{2}-2\sqrt{6}\right)\sqrt{2}}{4} and \frac{4\left(7+4\sqrt{3}\right)}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(6\sqrt{2}-2\sqrt{6}\right)\sqrt{2}-4\left(7+4\sqrt{3}\right)}{4}$$

Do the multiplications in \left(6\sqrt{2}-2\sqrt{6}\right)\sqrt{2}-4\left(7+4\sqrt{3}\right).

$$\frac{12-4\sqrt{3}-28-16\sqrt{3}}{4}$$

Do the calculations in 12-4\sqrt{3}-28-16\sqrt{3}.

$$\frac{-16-20\sqrt{3}}{4}$$

Divide each term of -16-20\sqrt{3} by 4 to get -4-5\sqrt{3}.

$$-4-5\sqrt{3}$$