Solve {left(sqrt{1/2}-1/2right)}^2

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

Evaluate

$$-\frac{\sqrt{2}}{2}+\frac{3}{4}\approx 0.042893219$$

Solution Steps

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

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

Calculate the square root of 1 and get 1.

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

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

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

The square of \sqrt{2} is 2.

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

Since \frac{\sqrt{2}}{2} and \frac{1}{2} have the same denominator, subtract them by subtracting their numerators.

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

To raise \frac{\sqrt{2}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}-1\right)^{2}.

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

$$\frac{3-2\sqrt{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{3-2\sqrt{2}}{4}$$

Expand

$$-\frac{\sqrt{2}}{2} + \frac{3}{4} = 0.042893219$$

Solution Steps

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

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

Calculate the square root of 1 and get 1.

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

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

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

The square of \sqrt{2} is 2.

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

Since \frac{\sqrt{2}}{2} and \frac{1}{2} have the same denominator, subtract them by subtracting their numerators.

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

To raise \frac{\sqrt{2}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}-1\right)^{2}.

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

$$\frac{3-2\sqrt{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{3-2\sqrt{2}}{4}$$