Solve (2(4/5+frac{2sqrt{199}}{5})-2)^2+(4/5+frac{2sqrt{99}}{5})^2

$$( 2 ( \frac { 4 } { 5 } + \frac { 2 \sqrt { 199 } } { 5 } ) - 2 ) ^ { 2 } + ( \frac { 4 } { 5 } + \frac { 2 \sqrt { 99 } } { 5 } ) ^ { 2 }$$

Evaluate

$$\frac{48\sqrt{11}}{25}-\frac{16\sqrt{199}}{25}+144\approx 141.33960857$$

Short Solution Steps

Since \frac{4}{5} and \frac{2\sqrt{199}}{5} have the same denominator, add them by adding their numerators.

$$\left(2\times \frac{4+2\sqrt{199}}{5}-2\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Express 2\times \frac{4+2\sqrt{199}}{5} as a single fraction.

$$\left(\frac{2\left(4+2\sqrt{199}\right)}{5}-2\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{5}{5}.

$$\left(\frac{2\left(4+2\sqrt{199}\right)}{5}-\frac{2\times 5}{5}\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Since \frac{2\left(4+2\sqrt{199}\right)}{5} and \frac{2\times 5}{5} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{2\left(4+2\sqrt{199}\right)-2\times 5}{5}\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Do the multiplications in 2\left(4+2\sqrt{199}\right)-2\times 5.

$$\left(\frac{8+4\sqrt{199}-10}{5}\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Do the calculations in 8+4\sqrt{199}-10.

$$\left(\frac{-2+4\sqrt{199}}{5}\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

To raise \frac{-2+4\sqrt{199}}{5} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Factor 99=3^{2}\times 11. Rewrite the square root of the product \sqrt{3^{2}\times 11} as the product of square roots \sqrt{3^{2}}\sqrt{11}. Take the square root of 3^{2}.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\left(\frac{4}{5}+\frac{2\times 3\sqrt{11}}{5}\right)^{2}$$

Multiply 2 and 3 to get 6.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\left(\frac{4}{5}+\frac{6\sqrt{11}}{5}\right)^{2}$$

Since \frac{4}{5} and \frac{6\sqrt{11}}{5} have the same denominator, add them by adding their numerators.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\left(\frac{4+6\sqrt{11}}{5}\right)^{2}$$

To raise \frac{4+6\sqrt{11}}{5} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Since \frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}} and \frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}+\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2+4\sqrt{199}\right)^{2}.

$$\frac{4-16\sqrt{199}+16\left(\sqrt{199}\right)^{2}}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

The square of \sqrt{199} is 199.

$$\frac{4-16\sqrt{199}+16\times 199}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Multiply 16 and 199 to get 3184.

$$\frac{4-16\sqrt{199}+3184}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Add 4 and 3184 to get 3188.

$$\frac{3188-16\sqrt{199}}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{3188-16\sqrt{199}}{25}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+6\sqrt{11}\right)^{2}.

$$\frac{3188-16\sqrt{199}}{25}+\frac{16+48\sqrt{11}+36\left(\sqrt{11}\right)^{2}}{5^{2}}$$

The square of \sqrt{11} is 11.

$$\frac{3188-16\sqrt{199}}{25}+\frac{16+48\sqrt{11}+36\times 11}{5^{2}}$$

Multiply 36 and 11 to get 396.

$$\frac{3188-16\sqrt{199}}{25}+\frac{16+48\sqrt{11}+396}{5^{2}}$$

Add 16 and 396 to get 412.

$$\frac{3188-16\sqrt{199}}{25}+\frac{412+48\sqrt{11}}{5^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{3188-16\sqrt{199}}{25}+\frac{412+48\sqrt{11}}{25}$$

Since \frac{3188-16\sqrt{199}}{25} and \frac{412+48\sqrt{11}}{25} have the same denominator, add them by adding their numerators.

$$\frac{3188-16\sqrt{199}+412+48\sqrt{11}}{25}$$

Do the calculations in 3188-16\sqrt{199}+412+48\sqrt{11}.

$$\frac{3600-16\sqrt{199}+48\sqrt{11}}{25}$$

Expand

$$\frac{48 \sqrt{11}}{25} - \frac{16 \sqrt{199}}{25} + 144 = 141.33960857$$

Short Solution Steps

Since \frac{4}{5} and \frac{2\sqrt{199}}{5} have the same denominator, add them by adding their numerators.

$$\left(2\times \frac{4+2\sqrt{199}}{5}-2\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Express 2\times \frac{4+2\sqrt{199}}{5} as a single fraction.

$$\left(\frac{2\left(4+2\sqrt{199}\right)}{5}-2\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{5}{5}.

$$\left(\frac{2\left(4+2\sqrt{199}\right)}{5}-\frac{2\times 5}{5}\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Since \frac{2\left(4+2\sqrt{199}\right)}{5} and \frac{2\times 5}{5} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{2\left(4+2\sqrt{199}\right)-2\times 5}{5}\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Do the multiplications in 2\left(4+2\sqrt{199}\right)-2\times 5.

$$\left(\frac{8+4\sqrt{199}-10}{5}\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Do the calculations in 8+4\sqrt{199}-10.

$$\left(\frac{-2+4\sqrt{199}}{5}\right)^{2}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

To raise \frac{-2+4\sqrt{199}}{5} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\left(\frac{4}{5}+\frac{2\sqrt{99}}{5}\right)^{2}$$

Factor 99=3^{2}\times 11. Rewrite the square root of the product \sqrt{3^{2}\times 11} as the product of square roots \sqrt{3^{2}}\sqrt{11}. Take the square root of 3^{2}.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\left(\frac{4}{5}+\frac{2\times 3\sqrt{11}}{5}\right)^{2}$$

Multiply 2 and 3 to get 6.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\left(\frac{4}{5}+\frac{6\sqrt{11}}{5}\right)^{2}$$

Since \frac{4}{5} and \frac{6\sqrt{11}}{5} have the same denominator, add them by adding their numerators.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\left(\frac{4+6\sqrt{11}}{5}\right)^{2}$$

To raise \frac{4+6\sqrt{11}}{5} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Since \frac{\left(-2+4\sqrt{199}\right)^{2}}{5^{2}} and \frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(-2+4\sqrt{199}\right)^{2}+\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2+4\sqrt{199}\right)^{2}.

$$\frac{4-16\sqrt{199}+16\left(\sqrt{199}\right)^{2}}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

The square of \sqrt{199} is 199.

$$\frac{4-16\sqrt{199}+16\times 199}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Multiply 16 and 199 to get 3184.

$$\frac{4-16\sqrt{199}+3184}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Add 4 and 3184 to get 3188.

$$\frac{3188-16\sqrt{199}}{5^{2}}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{3188-16\sqrt{199}}{25}+\frac{\left(4+6\sqrt{11}\right)^{2}}{5^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+6\sqrt{11}\right)^{2}.

$$\frac{3188-16\sqrt{199}}{25}+\frac{16+48\sqrt{11}+36\left(\sqrt{11}\right)^{2}}{5^{2}}$$

The square of \sqrt{11} is 11.

$$\frac{3188-16\sqrt{199}}{25}+\frac{16+48\sqrt{11}+36\times 11}{5^{2}}$$

Multiply 36 and 11 to get 396.

$$\frac{3188-16\sqrt{199}}{25}+\frac{16+48\sqrt{11}+396}{5^{2}}$$

Add 16 and 396 to get 412.

$$\frac{3188-16\sqrt{199}}{25}+\frac{412+48\sqrt{11}}{5^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{3188-16\sqrt{199}}{25}+\frac{412+48\sqrt{11}}{25}$$

Since \frac{3188-16\sqrt{199}}{25} and \frac{412+48\sqrt{11}}{25} have the same denominator, add them by adding their numerators.

$$\frac{3188-16\sqrt{199}+412+48\sqrt{11}}{25}$$

Do the calculations in 3188-16\sqrt{199}+412+48\sqrt{11}.

$$\frac{3600-16\sqrt{199}+48\sqrt{11}}{25}$$