Solve {l}{x^2-2xy+y^2=z}{x=3+sqrt{5}}{y=3-sqrt{5}}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { x ^ {2} - 2 x y + y ^ {2} = z }\\ { x = 3 + \sqrt{5} }\\ { y = 3 - \sqrt{5} }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b=20$$

Solution Steps

Consider the first equation. Insert the known values of variables into the equation.

$$\left(3+\sqrt{5}\right)^{2}-2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\left(3-\sqrt{5}\right)^{2}=z$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{5}\right)^{2}.

$$9+6\sqrt{5}+\left(\sqrt{5}\right)^{2}-2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\left(3-\sqrt{5}\right)^{2}=z$$

The square of \sqrt{5} is 5.

$$9+6\sqrt{5}+5-2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\left(3-\sqrt{5}\right)^{2}=z$$

Add 9 and 5 to get 14.

$$14+6\sqrt{5}-2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\left(3-\sqrt{5}\right)^{2}=z$$

Use the distributive property to multiply -2 by 3+\sqrt{5}.

$$14+6\sqrt{5}+\left(-6-2\sqrt{5}\right)\left(3-\sqrt{5}\right)+\left(3-\sqrt{5}\right)^{2}=z$$

Use the distributive property to multiply -6-2\sqrt{5} by 3-\sqrt{5} and combine like terms.

$$14+6\sqrt{5}-18+2\left(\sqrt{5}\right)^{2}+\left(3-\sqrt{5}\right)^{2}=z$$

The square of \sqrt{5} is 5.

$$14+6\sqrt{5}-18+2\times 5+\left(3-\sqrt{5}\right)^{2}=z$$

Multiply 2 and 5 to get 10.

$$14+6\sqrt{5}-18+10+\left(3-\sqrt{5}\right)^{2}=z$$

Add -18 and 10 to get -8.

$$14+6\sqrt{5}-8+\left(3-\sqrt{5}\right)^{2}=z$$

Subtract 8 from 14 to get 6.

$$6+6\sqrt{5}+\left(3-\sqrt{5}\right)^{2}=z$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-\sqrt{5}\right)^{2}.

$$6+6\sqrt{5}+9-6\sqrt{5}+\left(\sqrt{5}\right)^{2}=z$$

The square of \sqrt{5} is 5.

$$6+6\sqrt{5}+9-6\sqrt{5}+5=z$$

Add 9 and 5 to get 14.

$$6+6\sqrt{5}+14-6\sqrt{5}=z$$

Add 6 and 14 to get 20.

$$20+6\sqrt{5}-6\sqrt{5}=z$$

Combine 6\sqrt{5} and -6\sqrt{5} to get 0.

$$20=z$$

Swap sides so that all variable terms are on the left hand side.

$$z=20$$

Consider the fourth equation. Insert the known values of variables into the equation.

$$a=20$$

Consider the fifth equation. Insert the known values of variables into the equation.

$$b=20$$

The system is now solved.

$$x=3+\sqrt{5} y=3-\sqrt{5} z=20 a=20 b=20$$