Solve {left(2/5x-5/2right)}^2+{left(2/5-5/2xright)}^2=2{left(frac{2}{5}-frac{5}{2}xright)}^frac{2}{5}frac{5}{2}

$${ \left( \frac{ 2 }{ 5 } x- \frac{ 5 }{ 2 } \right) }^{ 2 } + { \left( \frac{ 2 }{ 5 } - \frac{ 5 }{ 2 } x \right) }^{ 2 } = 2 { \left( \frac{ 2 }{ 5 } - \frac{ 5 }{ 2 } x \right) }^{ } \frac{ 2 }{ 5 } \frac{ 5 }{ 2 }$$

Solve for x (complex solution)

$$x=\frac{-50+\sqrt{357101}i}{641}\approx -0.07800312+0.932260973i$$

$$x=\frac{-\sqrt{357101}i-50}{641}\approx -0.07800312-0.932260973i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2}{5}x-\frac{5}{2}\right)^{2}.

$$\frac{4}{25}x^{2}-2x+\frac{25}{4}+\left(\frac{2}{5}-\frac{5}{2}x\right)^{2}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2}{5}-\frac{5}{2}x\right)^{2}.

$$\frac{4}{25}x^{2}-2x+\frac{25}{4}+\frac{4}{25}-2x+\frac{25}{4}x^{2}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Add \frac{25}{4} and \frac{4}{25} to get \frac{641}{100}.

$$\frac{4}{25}x^{2}-2x+\frac{641}{100}-2x+\frac{25}{4}x^{2}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Combine -2x and -2x to get -4x.

$$\frac{4}{25}x^{2}-4x+\frac{641}{100}+\frac{25}{4}x^{2}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Combine \frac{4}{25}x^{2} and \frac{25}{4}x^{2} to get \frac{641}{100}x^{2}.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Calculate \frac{2}{5}-\frac{5}{2}x to the power of 1 and get \frac{2}{5}-\frac{5}{2}x.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=2\left(\frac{2}{5}-\frac{5}{2}x\right)\times \frac{2}{5}\times \frac{5}{2}$$

Multiply 2 and \frac{2}{5} to get \frac{4}{5}.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=\frac{4}{5}\left(\frac{2}{5}-\frac{5}{2}x\right)\times \frac{5}{2}$$

Multiply \frac{4}{5} and \frac{5}{2} to get 2.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=2\left(\frac{2}{5}-\frac{5}{2}x\right)$$

Use the distributive property to multiply 2 by \frac{2}{5}-\frac{5}{2}x.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=\frac{4}{5}-5x$$

Subtract \frac{4}{5} from both sides.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}-\frac{4}{5}=-5x$$

Subtract \frac{4}{5} from \frac{641}{100} to get \frac{561}{100}.

$$\frac{641}{100}x^{2}-4x+\frac{561}{100}=-5x$$

Add 5x to both sides.

$$\frac{641}{100}x^{2}-4x+\frac{561}{100}+5x=0$$

Combine -4x and 5x to get x.

$$\frac{641}{100}x^{2}+x+\frac{561}{100}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{641}{100} for a, 1 for b, and \frac{561}{100} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{641}{100}\times \frac{561}{100}}}{2\times \frac{641}{100}}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times \frac{641}{100}\times \frac{561}{100}}}{2\times \frac{641}{100}}$$

Multiply -4 times \frac{641}{100}.

$$x=\frac{-1±\sqrt{1-\frac{641}{25}\times \frac{561}{100}}}{2\times \frac{641}{100}}$$

Multiply -\frac{641}{25} times \frac{561}{100} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-1±\sqrt{1-\frac{359601}{2500}}}{2\times \frac{641}{100}}$$

Add 1 to -\frac{359601}{2500}.

$$x=\frac{-1±\sqrt{-\frac{357101}{2500}}}{2\times \frac{641}{100}}$$

Take the square root of -\frac{357101}{2500}.

$$x=\frac{-1±\frac{\sqrt{357101}i}{50}}{2\times \frac{641}{100}}$$

Multiply 2 times \frac{641}{100}.

$$x=\frac{-1±\frac{\sqrt{357101}i}{50}}{\frac{641}{50}}$$

Now solve the equation x=\frac{-1±\frac{\sqrt{357101}i}{50}}{\frac{641}{50}} when ± is plus. Add -1 to \frac{i\sqrt{357101}}{50}.

$$x=\frac{\frac{\sqrt{357101}i}{50}-1}{\frac{641}{50}}$$

Divide -1+\frac{i\sqrt{357101}}{50} by \frac{641}{50} by multiplying -1+\frac{i\sqrt{357101}}{50} by the reciprocal of \frac{641}{50}.

$$x=\frac{-50+\sqrt{357101}i}{641}$$

Now solve the equation x=\frac{-1±\frac{\sqrt{357101}i}{50}}{\frac{641}{50}} when ± is minus. Subtract \frac{i\sqrt{357101}}{50} from -1.

$$x=\frac{-\frac{\sqrt{357101}i}{50}-1}{\frac{641}{50}}$$

Divide -1-\frac{i\sqrt{357101}}{50} by \frac{641}{50} by multiplying -1-\frac{i\sqrt{357101}}{50} by the reciprocal of \frac{641}{50}.

$$x=\frac{-\sqrt{357101}i-50}{641}$$

The equation is now solved.

$$x=\frac{-50+\sqrt{357101}i}{641} x=\frac{-\sqrt{357101}i-50}{641}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2}{5}x-\frac{5}{2}\right)^{2}.

$$\frac{4}{25}x^{2}-2x+\frac{25}{4}+\left(\frac{2}{5}-\frac{5}{2}x\right)^{2}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2}{5}-\frac{5}{2}x\right)^{2}.

$$\frac{4}{25}x^{2}-2x+\frac{25}{4}+\frac{4}{25}-2x+\frac{25}{4}x^{2}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Add \frac{25}{4} and \frac{4}{25} to get \frac{641}{100}.

$$\frac{4}{25}x^{2}-2x+\frac{641}{100}-2x+\frac{25}{4}x^{2}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Combine -2x and -2x to get -4x.

$$\frac{4}{25}x^{2}-4x+\frac{641}{100}+\frac{25}{4}x^{2}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Combine \frac{4}{25}x^{2} and \frac{25}{4}x^{2} to get \frac{641}{100}x^{2}.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=2\left(\frac{2}{5}-\frac{5}{2}x\right)^{1}\times \frac{2}{5}\times \frac{5}{2}$$

Calculate \frac{2}{5}-\frac{5}{2}x to the power of 1 and get \frac{2}{5}-\frac{5}{2}x.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=2\left(\frac{2}{5}-\frac{5}{2}x\right)\times \frac{2}{5}\times \frac{5}{2}$$

Multiply 2 and \frac{2}{5} to get \frac{4}{5}.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=\frac{4}{5}\left(\frac{2}{5}-\frac{5}{2}x\right)\times \frac{5}{2}$$

Multiply \frac{4}{5} and \frac{5}{2} to get 2.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=2\left(\frac{2}{5}-\frac{5}{2}x\right)$$

Use the distributive property to multiply 2 by \frac{2}{5}-\frac{5}{2}x.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}=\frac{4}{5}-5x$$

Add 5x to both sides.

$$\frac{641}{100}x^{2}-4x+\frac{641}{100}+5x=\frac{4}{5}$$

Combine -4x and 5x to get x.

$$\frac{641}{100}x^{2}+x+\frac{641}{100}=\frac{4}{5}$$

Subtract \frac{641}{100} from both sides.

$$\frac{641}{100}x^{2}+x=\frac{4}{5}-\frac{641}{100}$$

Subtract \frac{641}{100} from \frac{4}{5} to get -\frac{561}{100}.

$$\frac{641}{100}x^{2}+x=-\frac{561}{100}$$

Divide both sides of the equation by \frac{641}{100}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{641}{100}x^{2}+x}{\frac{641}{100}}=-\frac{\frac{561}{100}}{\frac{641}{100}}$$

Dividing by \frac{641}{100} undoes the multiplication by \frac{641}{100}.

$$x^{2}+\frac{1}{\frac{641}{100}}x=-\frac{\frac{561}{100}}{\frac{641}{100}}$$

Divide 1 by \frac{641}{100} by multiplying 1 by the reciprocal of \frac{641}{100}.

$$x^{2}+\frac{100}{641}x=-\frac{\frac{561}{100}}{\frac{641}{100}}$$

Divide -\frac{561}{100} by \frac{641}{100} by multiplying -\frac{561}{100} by the reciprocal of \frac{641}{100}.

$$x^{2}+\frac{100}{641}x=-\frac{561}{641}$$

Divide \frac{100}{641}, the coefficient of the x term, by 2 to get \frac{50}{641}. Then add the square of \frac{50}{641} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{100}{641}x+\left(\frac{50}{641}\right)^{2}=-\frac{561}{641}+\left(\frac{50}{641}\right)^{2}$$

Square \frac{50}{641} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{100}{641}x+\frac{2500}{410881}=-\frac{561}{641}+\frac{2500}{410881}$$

Add -\frac{561}{641} to \frac{2500}{410881} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{100}{641}x+\frac{2500}{410881}=-\frac{357101}{410881}$$

Factor x^{2}+\frac{100}{641}x+\frac{2500}{410881}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{50}{641}\right)^{2}=-\frac{357101}{410881}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{50}{641}\right)^{2}}=\sqrt{-\frac{357101}{410881}}$$

Simplify.

$$x+\frac{50}{641}=\frac{\sqrt{357101}i}{641} x+\frac{50}{641}=-\frac{\sqrt{357101}i}{641}$$

Subtract \frac{50}{641} from both sides of the equation.

$$x=\frac{-50+\sqrt{357101}i}{641} x=\frac{-\sqrt{357101}i-50}{641}$$