Solve (frac{2sqrt{2}-m}{sqrt{2}})^2+[4-(frac{2sqrt{2}-m}{sqrt{2}})]^2=m^2+(2sqrt{2})^2

$$( \frac { 2 \sqrt { 2 } - m } { \sqrt { 2 } } ) ^ { 2 } + [ 4 - ( \frac { 2 \sqrt { 2 } - m } { \sqrt { 2 } } ) ] ^ { 2 } = m ^ { 2 } + ( 2 \sqrt { 2 } ) ^ { 2 }$$

Solve for m (complex solution)

$$m\in \mathrm{C}$$

Short Solution Steps

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

$$\left(\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)^{2}+\left(4-\frac{2\sqrt{2}-m}{\sqrt{2}}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\left(\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}+\left(4-\frac{2\sqrt{2}-m}{\sqrt{2}}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

To raise \frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+\left(4-\frac{2\sqrt{2}-m}{\sqrt{2}}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+\left(4-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+\left(4-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+16+8\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Use the distributive property to multiply 2\sqrt{2}-m by \sqrt{2}.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+16+8\left(-\frac{2\left(\sqrt{2}\right)^{2}-m\sqrt{2}}{2}\right)+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+16+8\left(-\frac{2\times 2-m\sqrt{2}}{2}\right)+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+16+8\left(-\frac{4-m\sqrt{2}}{2}\right)+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Cancel out 2, the greatest common factor in 8 and 2.

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

Use the distributive property to multiply -4 by 4-m\sqrt{2}.

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

Subtract 16 from 16 to get 0.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Use the distributive property to multiply 2\sqrt{2}-m by \sqrt{2}.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(-\frac{2\left(\sqrt{2}\right)^{2}-m\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(-\frac{2\times 2-m\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(-\frac{4-m\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Calculate -\frac{4-m\sqrt{2}}{2} to the power of 2 and get \left(\frac{4-m\sqrt{2}}{2}\right)^{2}.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(\frac{4-m\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4\sqrt{2}m times \frac{2^{2}}{2^{2}}.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+\frac{4\sqrt{2}m\times 2^{2}}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}+4\sqrt{2}m\times 2^{2}}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{16-8m\sqrt{2}+2m^{2}+16\sqrt{2}m}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Combine like terms in 16-8m\sqrt{2}+2m^{2}+16\sqrt{2}m.

$$\frac{16+8m\sqrt{2}+2m^{2}}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Factor the expressions that are not already factored in \frac{16+8m\sqrt{2}+2m^{2}}{2^{2}}.

$$\frac{2\left(m^{2}+4\sqrt{2}m+8\right)}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Cancel out 2 in both numerator and denominator.

$$\frac{m^{2}+4\sqrt{2}m+8}{2}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 2^{2} is 4. Multiply \frac{m^{2}+4\sqrt{2}m+8}{2} times \frac{2}{2}.

$$\frac{2\left(m^{2}+4\sqrt{2}m+8\right)}{4}+\frac{\left(4-m\sqrt{2}\right)^{2}}{4}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Since \frac{2\left(m^{2}+4\sqrt{2}m+8\right)}{4} and \frac{\left(4-m\sqrt{2}\right)^{2}}{4} have the same denominator, add them by adding their numerators.

$$\frac{2\left(m^{2}+4\sqrt{2}m+8\right)+\left(4-m\sqrt{2}\right)^{2}}{4}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Do the multiplications in 2\left(m^{2}+4\sqrt{2}m+8\right)+\left(4-m\sqrt{2}\right)^{2}.

$$\frac{2m^{2}+8\sqrt{2}m+16+16-8m\sqrt{2}+2m^{2}}{4}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Combine like terms in 2m^{2}+8\sqrt{2}m+16+16-8m\sqrt{2}+2m^{2}.

$$\frac{4m^{2}+32}{4}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Divide each term of 4m^{2}+32 by 4 to get m^{2}+8.

$$m^{2}+8=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Expand \left(2\sqrt{2}\right)^{2}.

$$m^{2}+8=m^{2}+2^{2}\left(\sqrt{2}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$m^{2}+8=m^{2}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$m^{2}+8=m^{2}+4\times 2$$

Multiply 4 and 2 to get 8.

$$m^{2}+8=m^{2}+8$$

Subtract m^{2} from both sides.

$$m^{2}+8-m^{2}=8$$

Combine m^{2} and -m^{2} to get 0.

$$8=8$$

Compare 8 and 8.

$$\text{true}$$

This is true for any m.

$$m\in \mathrm{C}$$

Solve for m

$$m\in \mathrm{R}$$

Short Solution Steps

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

$$\left(\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)^{2}+\left(4-\frac{2\sqrt{2}-m}{\sqrt{2}}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\left(\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}+\left(4-\frac{2\sqrt{2}-m}{\sqrt{2}}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

To raise \frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+\left(4-\frac{2\sqrt{2}-m}{\sqrt{2}}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+\left(4-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+\left(4-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+16+8\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Use the distributive property to multiply 2\sqrt{2}-m by \sqrt{2}.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+16+8\left(-\frac{2\left(\sqrt{2}\right)^{2}-m\sqrt{2}}{2}\right)+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+16+8\left(-\frac{2\times 2-m\sqrt{2}}{2}\right)+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+16+8\left(-\frac{4-m\sqrt{2}}{2}\right)+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Cancel out 2, the greatest common factor in 8 and 2.

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

Use the distributive property to multiply -4 by 4-m\sqrt{2}.

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

Subtract 16 from 16 to get 0.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(-\frac{\left(2\sqrt{2}-m\right)\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Use the distributive property to multiply 2\sqrt{2}-m by \sqrt{2}.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(-\frac{2\left(\sqrt{2}\right)^{2}-m\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(-\frac{2\times 2-m\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(-\frac{4-m\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Calculate -\frac{4-m\sqrt{2}}{2} to the power of 2 and get \left(\frac{4-m\sqrt{2}}{2}\right)^{2}.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\left(\frac{4-m\sqrt{2}}{2}\right)^{2}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+4\sqrt{2}m+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4\sqrt{2}m times \frac{2^{2}}{2^{2}}.

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}}{2^{2}}+\frac{4\sqrt{2}m\times 2^{2}}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{\left(\left(2\sqrt{2}-m\right)\sqrt{2}\right)^{2}+4\sqrt{2}m\times 2^{2}}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

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

$$\frac{16-8m\sqrt{2}+2m^{2}+16\sqrt{2}m}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Combine like terms in 16-8m\sqrt{2}+2m^{2}+16\sqrt{2}m.

$$\frac{16+8m\sqrt{2}+2m^{2}}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Factor the expressions that are not already factored in \frac{16+8m\sqrt{2}+2m^{2}}{2^{2}}.

$$\frac{2\left(m^{2}+4\sqrt{2}m+8\right)}{2^{2}}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Cancel out 2 in both numerator and denominator.

$$\frac{m^{2}+4\sqrt{2}m+8}{2}+\frac{\left(4-m\sqrt{2}\right)^{2}}{2^{2}}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 2^{2} is 4. Multiply \frac{m^{2}+4\sqrt{2}m+8}{2} times \frac{2}{2}.

$$\frac{2\left(m^{2}+4\sqrt{2}m+8\right)}{4}+\frac{\left(4-m\sqrt{2}\right)^{2}}{4}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Since \frac{2\left(m^{2}+4\sqrt{2}m+8\right)}{4} and \frac{\left(4-m\sqrt{2}\right)^{2}}{4} have the same denominator, add them by adding their numerators.

$$\frac{2\left(m^{2}+4\sqrt{2}m+8\right)+\left(4-m\sqrt{2}\right)^{2}}{4}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Do the multiplications in 2\left(m^{2}+4\sqrt{2}m+8\right)+\left(4-m\sqrt{2}\right)^{2}.

$$\frac{2m^{2}+8\sqrt{2}m+16+16-8m\sqrt{2}+2m^{2}}{4}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Combine like terms in 2m^{2}+8\sqrt{2}m+16+16-8m\sqrt{2}+2m^{2}.

$$\frac{4m^{2}+32}{4}=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Divide each term of 4m^{2}+32 by 4 to get m^{2}+8.

$$m^{2}+8=m^{2}+\left(2\sqrt{2}\right)^{2}$$

Expand \left(2\sqrt{2}\right)^{2}.

$$m^{2}+8=m^{2}+2^{2}\left(\sqrt{2}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$m^{2}+8=m^{2}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$m^{2}+8=m^{2}+4\times 2$$

Multiply 4 and 2 to get 8.

$$m^{2}+8=m^{2}+8$$

Subtract m^{2} from both sides.

$$m^{2}+8-m^{2}=8$$

Combine m^{2} and -m^{2} to get 0.

$$8=8$$

Compare 8 and 8.

$$\text{true}$$

This is true for any m.

$$m\in \mathrm{R}$$