計算第2題：
令 \(\displaystyle y=\log_4\left(x^2\right)\Rightarrow y=\log_2\left|x\right|\)，
\(\displaystyle x+\sqrt{12+8\left(\log_4 x^2\right)-4\left(\log_4 x^2\right)^2}=1\)
\(\displaystyle \Rightarrow x+\sqrt{12+8y-4y^2}=1\)
\(\displaystyle \Rightarrow \sqrt{12+8y-4y^2}=1-x\)
\(\displaystyle \Rightarrow 12+8y-4y^2=\left(1-x\right)^2\) 且 \(1-x\geq0\)
\(\displaystyle \Rightarrow \left(x-1\right)^2+4\left(y-1\right)^2=16\) 且 \(x\leq1\)
\(\displaystyle \Rightarrow \frac{\left(x-1\right)^2}{16}+\frac{\left(y-1\right)^2}{4}=1\) 且 \(x\leq1\)
作橢圓與 \(\displaystyle y=\log_2\left|x\right|\) 的圖形如下：
可知在 \(x\leq 1\) 的部分共有三個相異交點，
對應到原方程式有三個相異實根。
