As Easy As Phi

• $\frac{A+B}{A} = X$

  $A+B = AX$

  $B = AX - A$

• $\frac{A}{ B} = X$

  $B = \frac{A}{X}$

  $AX-A= \frac{A}{X}$

  $AX^2 - AX = A$

  $AX^2 -AX - A = 0$

  $A(X^2 - X - 1) = 0$

Since $A \neq 0$ , hence:

$X^2 - X - 1 = 0$

Solve for $X$ we get:

$\boxed{X = 0.618}$ or $X = -0.618$

$AB+B^2-A^2=0...solving for B, B=\dfrac{-A \pm \sqrt{A^2+4A^2}}{2} \\\therefore X=\dfrac{A}{B}= \dfrac{2}{1\pm \sqrt5} =1.618~~ or~~ -.618.$

$\frac{A+B}{A}=\frac{A}{B}$

${A}^{2} - AB - {B}^{2} = 0$

${({A}^{2} - \frac{B}{2})}^{2} = \frac{5{B}^{2}}{2}$

$A - \frac{B}{2}\ = B\frac{\sqrt{5}}{2}$

$\frac{A}{B}=\frac{1+\sqrt{5}}{2}$

${X}=\frac{A}{B}=\frac{1+\sqrt{5}}{2}$

$X\approx 1.618$

$\frac{a+b}{a}=1+\frac{b}{a}=1+(\frac{a}{b})^{-1}=\frac{a}{b}$

$(\frac{a}{b})^{2}-\frac{a}{b}-1=0$

x = a/b = $\frac{1\pm\sqrt{5}}{2}$ , reject the negative solution because that implies one of a or b are negative

We apply this rule

$\frac { a }{ b } =\frac { c }{ d } =\frac { a+c }{ b+d } \quad \$

Substituting $a = A+B, b = A, c = A, d = B$

$\frac { A+B }{ A } =\frac { A }{ B } =\frac { 2A+B }{ A+B } =1+\frac { A }{ A+B } =1+\frac { 1 }{ X } = X$

$\Leftrightarrow { X }^{ 2 }-X-1=0$

Solving for X, taking only the positive root as $A, B > 0$

$\Rightarrow X=\frac { 1+\sqrt { 5 } }{ 2 }$

$X\approx 1.618$

(x+1)/x = x, so we have quadratic eqs. x^2 - x - 1 = 0, and the solution is x=1.618