แก้โจทย์ {l}{x^{2}-ab-a^{2}=bx}{x^2=2a^2b^2+abx}{ax^2=a+b-bx}

$$\left. \begin{array} { l } { x ^ { 2 } - a b - a ^ { 2 } = b x } \\ { x ^ { 2 } = 2 a ^ { 2 } b ^ { 2 } + a b x } \\ { a x ^ { 2 } = a + b - b x } \end{array} \right.$$

หาค่า x, a, b

$$x=0\text{, }a=0\text{, }b=0$$

$$x=1\text{, }a=\frac{\sqrt{5}+1}{2}\approx 1.618033989\text{, }b=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

$$x=1\text{, }a=\frac{1-\sqrt{5}}{2}\approx -0.618033989\text{, }b=\frac{\sqrt{5}+1}{2}\approx 1.618033989$$

$$x=-\frac{2}{3}\approx -0.666666667\text{, }a=-1\text{, }b=\frac{1}{3}\approx 0.333333333$$

$$x=1\text{, }a=-1\text{, }b=1$$

$$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989\text{, }a=\frac{\sqrt{5}+1}{2}\approx 1.618033989\text{, }b=1$$

$$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989\text{, }a=\frac{1-\sqrt{5}}{2}\approx -0.618033989\text{, }b=1$$

$$x=1\text{, }a=-1\text{, }b=-\frac{1}{2}=-0.5$$