Solve 2(1/2-frac{sqrt{6}}{4})^2-1

$$2 ( \frac { 1 } { 2 } - \frac { \sqrt { 6 } } { 4 } ) ^ { 2 } - 1$$

Evaluate

$$-\frac{\sqrt{6}}{2}+\frac{1}{4}\approx -0.974744871$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 4 is 4. Multiply \frac{1}{2} times \frac{2}{2}.

$$2\left(\frac{2}{4}-\frac{\sqrt{6}}{4}\right)^{2}-1$$

Since \frac{2}{4} and \frac{\sqrt{6}}{4} have the same denominator, subtract them by subtracting their numerators.

$$2\times \left(\frac{2-\sqrt{6}}{4}\right)^{2}-1$$

To raise \frac{2-\sqrt{6}}{4} to a power, raise both numerator and denominator to the power and then divide.

$$2\times \frac{\left(2-\sqrt{6}\right)^{2}}{4^{2}}-1$$

Express 2\times \frac{\left(2-\sqrt{6}\right)^{2}}{4^{2}} as a single fraction.

$$\frac{2\left(2-\sqrt{6}\right)^{2}}{4^{2}}-1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{4^{2}}{4^{2}}.

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

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

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

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

$$\frac{2\left(4-4\sqrt{6}+\left(\sqrt{6}\right)^{2}\right)}{4^{2}}-1$$

The square of \sqrt{6} is 6.

$$\frac{2\left(4-4\sqrt{6}+6\right)}{4^{2}}-1$$

Add 4 and 6 to get 10.

$$\frac{2\left(10-4\sqrt{6}\right)}{4^{2}}-1$$

Calculate 4 to the power of 2 and get 16.

$$\frac{2\left(10-4\sqrt{6}\right)}{16}-1$$

Divide 2\left(10-4\sqrt{6}\right) by 16 to get \frac{1}{8}\left(10-4\sqrt{6}\right).

$$\frac{1}{8}\left(10-4\sqrt{6}\right)-1$$

Use the distributive property to multiply \frac{1}{8} by 10-4\sqrt{6}.

$$\frac{5}{4}-\frac{1}{2}\sqrt{6}-1$$

Subtract 1 from \frac{5}{4} to get \frac{1}{4}.

$$\frac{1}{4}-\frac{1}{2}\sqrt{6}$$

Expand

$$-\frac{\sqrt{6}}{2} + \frac{1}{4} = -0.974744871$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 4 is 4. Multiply \frac{1}{2} times \frac{2}{2}.

$$2\left(\frac{2}{4}-\frac{\sqrt{6}}{4}\right)^{2}-1$$

Since \frac{2}{4} and \frac{\sqrt{6}}{4} have the same denominator, subtract them by subtracting their numerators.

$$2\times \left(\frac{2-\sqrt{6}}{4}\right)^{2}-1$$

To raise \frac{2-\sqrt{6}}{4} to a power, raise both numerator and denominator to the power and then divide.

$$2\times \frac{\left(2-\sqrt{6}\right)^{2}}{4^{2}}-1$$

Express 2\times \frac{\left(2-\sqrt{6}\right)^{2}}{4^{2}} as a single fraction.

$$\frac{2\left(2-\sqrt{6}\right)^{2}}{4^{2}}-1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{4^{2}}{4^{2}}.

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

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

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

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

$$\frac{2\left(4-4\sqrt{6}+\left(\sqrt{6}\right)^{2}\right)}{4^{2}}-1$$

The square of \sqrt{6} is 6.

$$\frac{2\left(4-4\sqrt{6}+6\right)}{4^{2}}-1$$

Add 4 and 6 to get 10.

$$\frac{2\left(10-4\sqrt{6}\right)}{4^{2}}-1$$

Calculate 4 to the power of 2 and get 16.

$$\frac{2\left(10-4\sqrt{6}\right)}{16}-1$$

Divide 2\left(10-4\sqrt{6}\right) by 16 to get \frac{1}{8}\left(10-4\sqrt{6}\right).

$$\frac{1}{8}\left(10-4\sqrt{6}\right)-1$$

Use the distributive property to multiply \frac{1}{8} by 10-4\sqrt{6}.

$$\frac{5}{4}-\frac{1}{2}\sqrt{6}-1$$

Subtract 1 from \frac{5}{4} to get \frac{1}{4}.

$$\frac{1}{4}-\frac{1}{2}\sqrt{6}$$