Solve (frac{sqrt{73}}{2}-sqrt{10})^2

$$( \frac { \sqrt { 73 } } { 2 } - \sqrt { 10 } ) ^ { 2 }$$

Evaluate

$$\frac{113}{4}-\sqrt{730}\approx 1.231487828$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{10} times \frac{2}{2}.

$$\left(\frac{\sqrt{73}}{2}-\frac{2\sqrt{10}}{2}\right)^{2}$$

Since \frac{\sqrt{73}}{2} and \frac{2\sqrt{10}}{2} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{\sqrt{73}-2\sqrt{10}}{2}\right)^{2}$$

To raise \frac{\sqrt{73}-2\sqrt{10}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{73}-2\sqrt{10}\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{73}-2\sqrt{10}\right)^{2}.

$$\frac{\left(\sqrt{73}\right)^{2}-4\sqrt{73}\sqrt{10}+4\left(\sqrt{10}\right)^{2}}{2^{2}}$$

The square of \sqrt{73} is 73.

$$\frac{73-4\sqrt{73}\sqrt{10}+4\left(\sqrt{10}\right)^{2}}{2^{2}}$$

To multiply \sqrt{73} and \sqrt{10}, multiply the numbers under the square root.

$$\frac{73-4\sqrt{730}+4\left(\sqrt{10}\right)^{2}}{2^{2}}$$

The square of \sqrt{10} is 10.

$$\frac{73-4\sqrt{730}+4\times 10}{2^{2}}$$

Multiply 4 and 10 to get 40.

$$\frac{73-4\sqrt{730}+40}{2^{2}}$$

Add 73 and 40 to get 113.

$$\frac{113-4\sqrt{730}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{113-4\sqrt{730}}{4}$$

Expand

$$\frac{113}{4} - \sqrt{730} = 1.231487828$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{10} times \frac{2}{2}.

$$\left(\frac{\sqrt{73}}{2}-\frac{2\sqrt{10}}{2}\right)^{2}$$

Since \frac{\sqrt{73}}{2} and \frac{2\sqrt{10}}{2} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{\sqrt{73}-2\sqrt{10}}{2}\right)^{2}$$

To raise \frac{\sqrt{73}-2\sqrt{10}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{73}-2\sqrt{10}\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{73}-2\sqrt{10}\right)^{2}.

$$\frac{\left(\sqrt{73}\right)^{2}-4\sqrt{73}\sqrt{10}+4\left(\sqrt{10}\right)^{2}}{2^{2}}$$

The square of \sqrt{73} is 73.

$$\frac{73-4\sqrt{73}\sqrt{10}+4\left(\sqrt{10}\right)^{2}}{2^{2}}$$

To multiply \sqrt{73} and \sqrt{10}, multiply the numbers under the square root.

$$\frac{73-4\sqrt{730}+4\left(\sqrt{10}\right)^{2}}{2^{2}}$$

The square of \sqrt{10} is 10.

$$\frac{73-4\sqrt{730}+4\times 10}{2^{2}}$$

Multiply 4 and 10 to get 40.

$$\frac{73-4\sqrt{730}+40}{2^{2}}$$

Add 73 and 40 to get 113.

$$\frac{113-4\sqrt{730}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{113-4\sqrt{730}}{4}$$