Solve (frac{sqrt{41}+3}{2})^2+(sqrt{41}-3)^2-41-3/2

$$( \frac { \sqrt { 41 } + 3 } { 2 } ) ^ { 2 } + ( \sqrt { 41 } - 3 ) ^ { 2 } - \frac { 41 - 3 } { 2 }$$

Evaluate

$$\frac{87-9\sqrt{41}}{2}\approx 14.685940932$$

Short Solution Steps

To raise \frac{\sqrt{41}+3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+\left(\sqrt{41}-3\right)^{2}-\frac{41-3}{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{41}-3\right)^{2}.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+\left(\sqrt{41}\right)^{2}-6\sqrt{41}+9-\frac{41-3}{2}$$

The square of \sqrt{41} is 41.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+41-6\sqrt{41}+9-\frac{41-3}{2}$$

Add 41 and 9 to get 50.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+50-6\sqrt{41}-\frac{41-3}{2}$$

Subtract 3 from 41 to get 38.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+50-6\sqrt{41}-\frac{38}{2}$$

Divide 38 by 2 to get 19.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+50-6\sqrt{41}-19$$

Subtract 19 from 50 to get 31.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+31-6\sqrt{41}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 31-6\sqrt{41} times \frac{2^{2}}{2^{2}}.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+\frac{\left(31-6\sqrt{41}\right)\times 2^{2}}{2^{2}}$$

Since \frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}} and \frac{\left(31-6\sqrt{41}\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(\sqrt{41}+3\right)^{2}+\left(31-6\sqrt{41}\right)\times 2^{2}}{2^{2}}$$

Do the multiplications in \left(\sqrt{41}+3\right)^{2}+\left(31-6\sqrt{41}\right)\times 2^{2}.

$$\frac{\left(\sqrt{41}\right)^{2}+6\sqrt{41}+9+124-24\sqrt{41}}{2^{2}}$$

Do the calculations in \left(\sqrt{41}\right)^{2}+6\sqrt{41}+9+124-24\sqrt{41}.

$$\frac{174-18\sqrt{41}}{2^{2}}$$

Expand 2^{2}.

$$\frac{174-18\sqrt{41}}{4}$$

Expand

$$\frac{87 - 9 \sqrt{41}}{2} = 14.685940931552182$$

Short Solution Steps

To raise \frac{\sqrt{41}+3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+\left(\sqrt{41}-3\right)^{2}-\frac{41-3}{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{41}-3\right)^{2}.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+\left(\sqrt{41}\right)^{2}-6\sqrt{41}+9-\frac{41-3}{2}$$

The square of \sqrt{41} is 41.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+41-6\sqrt{41}+9-\frac{41-3}{2}$$

Add 41 and 9 to get 50.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+50-6\sqrt{41}-\frac{41-3}{2}$$

Subtract 3 from 41 to get 38.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+50-6\sqrt{41}-\frac{38}{2}$$

Divide 38 by 2 to get 19.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+50-6\sqrt{41}-19$$

Subtract 19 from 50 to get 31.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+31-6\sqrt{41}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 31-6\sqrt{41} times \frac{2^{2}}{2^{2}}.

$$\frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}}+\frac{\left(31-6\sqrt{41}\right)\times 2^{2}}{2^{2}}$$

Since \frac{\left(\sqrt{41}+3\right)^{2}}{2^{2}} and \frac{\left(31-6\sqrt{41}\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(\sqrt{41}+3\right)^{2}+\left(31-6\sqrt{41}\right)\times 2^{2}}{2^{2}}$$

Do the multiplications in \left(\sqrt{41}+3\right)^{2}+\left(31-6\sqrt{41}\right)\times 2^{2}.

$$\frac{\left(\sqrt{41}\right)^{2}+6\sqrt{41}+9+124-24\sqrt{41}}{2^{2}}$$

Do the calculations in \left(\sqrt{41}\right)^{2}+6\sqrt{41}+9+124-24\sqrt{41}.

$$\frac{174-18\sqrt{41}}{2^{2}}$$

Expand 2^{2}.

$$\frac{174-18\sqrt{41}}{4}$$