Solve 2sqrt{24}-8divsqrt{2/3}+(sqrt{3}-sqrt{2})^2

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

Evaluate

$$5-2\sqrt{6}\approx 0.101020514$$

Solution Steps

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}.

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

Multiply 2 and 2 to get 4.

$$4\sqrt{6}-\frac{8}{\sqrt{\frac{2}{3}}}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Rewrite the square root of the division \sqrt{\frac{2}{3}} as the division of square roots \frac{\sqrt{2}}{\sqrt{3}}.

$$4\sqrt{6}-\frac{8}{\frac{\sqrt{2}}{\sqrt{3}}}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Rationalize the denominator of \frac{\sqrt{2}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$4\sqrt{6}-\frac{8}{\frac{\sqrt{2}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$4\sqrt{6}-\frac{8}{\frac{\sqrt{2}\sqrt{3}}{3}}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$4\sqrt{6}-\frac{8}{\frac{\sqrt{6}}{3}}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Divide 8 by \frac{\sqrt{6}}{3} by multiplying 8 by the reciprocal of \frac{\sqrt{6}}{3}.

$$4\sqrt{6}-\frac{8\times 3}{\sqrt{6}}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Rationalize the denominator of \frac{8\times 3}{\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.

$$4\sqrt{6}-\frac{8\times 3\sqrt{6}}{\left(\sqrt{6}\right)^{2}}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

The square of \sqrt{6} is 6.

$$4\sqrt{6}-\frac{8\times 3\sqrt{6}}{6}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Multiply 8 and 3 to get 24.

$$4\sqrt{6}-\frac{24\sqrt{6}}{6}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Divide 24\sqrt{6} by 6 to get 4\sqrt{6}.

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

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

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

The square of \sqrt{3} is 3.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

The square of \sqrt{2} is 2.

$$4\sqrt{6}-4\sqrt{6}+3-2\sqrt{6}+2$$

Add 3 and 2 to get 5.

$$4\sqrt{6}-4\sqrt{6}+5-2\sqrt{6}$$

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

$$2\sqrt{6}-4\sqrt{6}+5$$

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

$$-2\sqrt{6}+5$$