Solve (frac{68-4sqrt{17}}{17})^2+4^2

$$( \frac { 68 - 4 \sqrt { 17 } } { 17 } ) ^ { 2 } + 4 ^ { 2 }$$

Evaluate

$$\frac{560-32\sqrt{17}}{17}\approx 25.180036469$$

Short Solution Steps

To raise \frac{68-4\sqrt{17}}{17} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(68-4\sqrt{17}\right)^{2}}{17^{2}}+4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(68-4\sqrt{17}\right)^{2}}{17^{2}}+16$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 16 times \frac{17^{2}}{17^{2}}.

$$\frac{\left(68-4\sqrt{17}\right)^{2}}{17^{2}}+\frac{16\times 17^{2}}{17^{2}}$$

Since \frac{\left(68-4\sqrt{17}\right)^{2}}{17^{2}} and \frac{16\times 17^{2}}{17^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(68-4\sqrt{17}\right)^{2}+16\times 17^{2}}{17^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(68-4\sqrt{17}\right)^{2}.

$$\frac{4624-544\sqrt{17}+16\left(\sqrt{17}\right)^{2}}{17^{2}}+16$$

The square of \sqrt{17} is 17.

$$\frac{4624-544\sqrt{17}+16\times 17}{17^{2}}+16$$

Multiply 16 and 17 to get 272.

$$\frac{4624-544\sqrt{17}+272}{17^{2}}+16$$

Add 4624 and 272 to get 4896.

$$\frac{4896-544\sqrt{17}}{17^{2}}+16$$

Calculate 17 to the power of 2 and get 289.

$$\frac{4896-544\sqrt{17}}{289}+16$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 16 times \frac{289}{289}.

$$\frac{4896-544\sqrt{17}}{289}+\frac{16\times 289}{289}$$

Since \frac{4896-544\sqrt{17}}{289} and \frac{16\times 289}{289} have the same denominator, add them by adding their numerators.

$$\frac{4896-544\sqrt{17}+16\times 289}{289}$$

Do the multiplications in 4896-544\sqrt{17}+16\times 289.

$$\frac{4896-544\sqrt{17}+4624}{289}$$

Do the calculations in 4896-544\sqrt{17}+4624.

$$\frac{9520-544\sqrt{17}}{289}$$

Expand

$$\frac{560 - 32 \sqrt{17}}{17} = 25.180036469425577$$

Short Solution Steps

To raise \frac{68-4\sqrt{17}}{17} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(68-4\sqrt{17}\right)^{2}}{17^{2}}+4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(68-4\sqrt{17}\right)^{2}}{17^{2}}+16$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 16 times \frac{17^{2}}{17^{2}}.

$$\frac{\left(68-4\sqrt{17}\right)^{2}}{17^{2}}+\frac{16\times 17^{2}}{17^{2}}$$

Since \frac{\left(68-4\sqrt{17}\right)^{2}}{17^{2}} and \frac{16\times 17^{2}}{17^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(68-4\sqrt{17}\right)^{2}+16\times 17^{2}}{17^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(68-4\sqrt{17}\right)^{2}.

$$\frac{4624-544\sqrt{17}+16\left(\sqrt{17}\right)^{2}}{17^{2}}+16$$

The square of \sqrt{17} is 17.

$$\frac{4624-544\sqrt{17}+16\times 17}{17^{2}}+16$$

Multiply 16 and 17 to get 272.

$$\frac{4624-544\sqrt{17}+272}{17^{2}}+16$$

Add 4624 and 272 to get 4896.

$$\frac{4896-544\sqrt{17}}{17^{2}}+16$$

Calculate 17 to the power of 2 and get 289.

$$\frac{4896-544\sqrt{17}}{289}+16$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 16 times \frac{289}{289}.

$$\frac{4896-544\sqrt{17}}{289}+\frac{16\times 289}{289}$$

Since \frac{4896-544\sqrt{17}}{289} and \frac{16\times 289}{289} have the same denominator, add them by adding their numerators.

$$\frac{4896-544\sqrt{17}+16\times 289}{289}$$

Do the multiplications in 4896-544\sqrt{17}+16\times 289.

$$\frac{4896-544\sqrt{17}+4624}{289}$$

Do the calculations in 4896-544\sqrt{17}+4624.

$$\frac{9520-544\sqrt{17}}{289}$$