แก้โจทย์ 9x^2-6b^2x-(a^4-b^4)=0

$$9 x ^ { 2 } - 6 b ^ { 2 } x - ( a ^ { 4 } - b ^ { 4 } ) = 0$$

หาค่า a (complex solution)

$$a=-\sqrt{b^{2}-3x}$$

$$a=\sqrt{b^{2}-3x}$$

$$a=\sqrt{3x-b^{2}}$$

$$a=-\sqrt{3x-b^{2}}$$

หาค่า b (complex solution)

$$b=\sqrt{3x-a^{2}}$$

$$b=-\sqrt{3x-a^{2}}$$

$$b=-\sqrt{3x+a^{2}}$$

$$b=\sqrt{3x+a^{2}}$$

หาค่า a

$$\left\{\begin{matrix}a=\sqrt{b^{2}-3x}\text{; }a=-\sqrt{b^{2}-3x}\text{, }&x\leq \frac{b^{2}}{3}\\a=\sqrt{3x-b^{2}}\text{; }a=-\sqrt{3x-b^{2}}\text{, }&x\geq \frac{b^{2}}{3}\end{matrix}\right.$$

หาค่า b

$$\left\{\begin{matrix}b=-\sqrt{3x-a^{2}}\text{; }b=\sqrt{3x-a^{2}}\text{, }&x\geq \frac{a^{2}}{3}\\b=-\sqrt{3x+a^{2}}\text{; }b=\sqrt{3x+a^{2}}\text{, }&x\geq -\frac{a^{2}}{3}\end{matrix}\right.$$