Solve [(1/2)^4+(1/2)^2]-3[(1/sqrt{2})^2-1]-(frac{sqrt{3}}{2})

$$[ ( \frac { 1 } { 2 } ) ^ { 4 } + ( \frac { 1 } { 2 } ) ^ { 2 } ] - 3 [ ( \frac { 1 } { \sqrt { 2 } } ) ^ { 2 } - 1 ] - ( \frac { \sqrt { 3 } } { 2 } )$$

Evaluate

$$-\frac{\sqrt{3}}{2}+\frac{29}{16}\approx 0.946474596$$

Short Solution Steps

Calculate \frac{1}{2} to the power of 4 and get \frac{1}{16}.

$$\frac{1}{16}+\left(\frac{1}{2}\right)^{2}-3\left(\left(\frac{1}{\sqrt{2}}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{16}+\frac{1}{4}-3\left(\left(\frac{1}{\sqrt{2}}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

Add \frac{1}{16} and \frac{1}{4} to get \frac{5}{16}.

$$\frac{5}{16}-3\left(\left(\frac{1}{\sqrt{2}}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

Rationalize the denominator of \frac{1}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{5}{16}-3\left(\left(\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

The square of \sqrt{2} is 2.

$$\frac{5}{16}-3\left(\left(\frac{\sqrt{2}}{2}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

To raise \frac{\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{5}{16}-3\left(\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-1\right)-\frac{\sqrt{3}}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2^{2}}{2^{2}}.

$$\frac{5}{16}-3\left(\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-\frac{2^{2}}{2^{2}}\right)-\frac{\sqrt{3}}{2}$$

Since \frac{\left(\sqrt{2}\right)^{2}}{2^{2}} and \frac{2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5}{16}-3\times \frac{\left(\sqrt{2}\right)^{2}-2^{2}}{2^{2}}-\frac{\sqrt{3}}{2}$$

Express 3\times \frac{\left(\sqrt{2}\right)^{2}-2^{2}}{2^{2}} as a single fraction.

$$\frac{5}{16}-\frac{3\left(\left(\sqrt{2}\right)^{2}-2^{2}\right)}{2^{2}}-\frac{\sqrt{3}}{2}$$

The square of \sqrt{2} is 2.

$$\frac{5}{16}-\frac{3\left(2-2^{2}\right)}{2^{2}}-\frac{\sqrt{3}}{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{5}{16}-\frac{3\left(2-4\right)}{2^{2}}-\frac{\sqrt{3}}{2}$$

Subtract 4 from 2 to get -2.

$$\frac{5}{16}-\frac{3\left(-2\right)}{2^{2}}-\frac{\sqrt{3}}{2}$$

Multiply 3 and -2 to get -6.

$$\frac{5}{16}-\frac{-6}{2^{2}}-\frac{\sqrt{3}}{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{5}{16}-\frac{-6}{4}-\frac{\sqrt{3}}{2}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$\frac{5}{16}-\left(-\frac{3}{2}\right)-\frac{\sqrt{3}}{2}$$

The opposite of -\frac{3}{2} is \frac{3}{2}.

$$\frac{5}{16}+\frac{3}{2}-\frac{\sqrt{3}}{2}$$

Add \frac{5}{16} and \frac{3}{2} to get \frac{29}{16}.

$$\frac{29}{16}-\frac{\sqrt{3}}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 16 and 2 is 16. Multiply \frac{\sqrt{3}}{2} times \frac{8}{8}.

$$\frac{29}{16}-\frac{8\sqrt{3}}{16}$$

Since \frac{29}{16} and \frac{8\sqrt{3}}{16} have the same denominator, subtract them by subtracting their numerators.

$$\frac{29-8\sqrt{3}}{16}$$

Factor

$$\frac{29 - 8 \sqrt{3}}{16} = 0.9464745962155614$$