Solve sqrt{8}-1/8sqrt{48}-(2/3sqrt{41/2}-2sqrt{3/4})

$$\sqrt { 8 } - \frac { 1 } { 8 } \sqrt { 48 } - ( \frac { 2 } { 3 } \sqrt { 4 \frac { 1 } { 2 } } - 2 \sqrt { \frac { 3 } { 4 } } )$$

Evaluate

$$\frac{\sqrt{3}}{2}+\sqrt{2}\approx 2.280238966$$

Solution Steps

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}.

$$2\sqrt{2}-\frac{1}{8}\sqrt{48}-\left(\frac{2}{3}\sqrt{\frac{4\times 2+1}{2}}-2\sqrt{\frac{3}{4}}\right)$$

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}.

$$2\sqrt{2}-\frac{1}{8}\times 4\sqrt{3}-\left(\frac{2}{3}\sqrt{\frac{4\times 2+1}{2}}-2\sqrt{\frac{3}{4}}\right)$$

Express -\frac{1}{8}\times 4 as a single fraction.

$$2\sqrt{2}+\frac{-4}{8}\sqrt{3}-\left(\frac{2}{3}\sqrt{\frac{4\times 2+1}{2}}-2\sqrt{\frac{3}{4}}\right)$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

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

Multiply 4 and 2 to get 8.

$$2\sqrt{2}-\frac{1}{2}\sqrt{3}-\left(\frac{2}{3}\sqrt{\frac{8+1}{2}}-2\sqrt{\frac{3}{4}}\right)$$

Add 8 and 1 to get 9.

$$2\sqrt{2}-\frac{1}{2}\sqrt{3}-\left(\frac{2}{3}\sqrt{\frac{9}{2}}-2\sqrt{\frac{3}{4}}\right)$$

Rewrite the square root of the division \sqrt{\frac{9}{2}} as the division of square roots \frac{\sqrt{9}}{\sqrt{2}}.

$$2\sqrt{2}-\frac{1}{2}\sqrt{3}-\left(\frac{2}{3}\times \frac{\sqrt{9}}{\sqrt{2}}-2\sqrt{\frac{3}{4}}\right)$$

Calculate the square root of 9 and get 3.

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

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

$$2\sqrt{2}-\frac{1}{2}\sqrt{3}-\left(\frac{2}{3}\times \frac{3\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-2\sqrt{\frac{3}{4}}\right)$$

The square of \sqrt{2} is 2.

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

Multiply \frac{2}{3} times \frac{3\sqrt{2}}{2} by multiplying numerator times numerator and denominator times denominator.

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

Cancel out 2\times 3 in both numerator and denominator.

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

Rewrite the square root of the division \sqrt{\frac{3}{4}} as the division of square roots \frac{\sqrt{3}}{\sqrt{4}}.

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

Calculate the square root of 4 and get 2.

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

Cancel out 2 and 2.

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

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

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

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

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

The opposite of -\sqrt{3} is \sqrt{3}.

$$\sqrt{2}-\frac{1}{2}\sqrt{3}+\sqrt{3}$$

Combine -\frac{1}{2}\sqrt{3} and \sqrt{3} to get \frac{1}{2}\sqrt{3}.

$$\sqrt{2}+\frac{1}{2}\sqrt{3}$$

Factor

$$\frac{\sqrt{3} + 2 \sqrt{2}}{2} = 2.2802389661575337$$