ข้อสอบคณิต 2 - มีนาคม 2562

$$\begin{gathered} \mathrm{จากโจทย์} \\ 4^{x+\frac{1}{2}}+2^x-1 = 0 \\ {\left(2^2\right)}^{x+\frac{1}{2}}+2^x-1 = 0 \\ 2^{2x+1}+2^x-1 = 0 \\ 2^{2x}·2+2^x-1 = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{ให้} A = 2^x \\ 2A^2+A-1 = 0 \\ \left(2A-1\right)\left(A+1\right) = 0 \\ A = \frac{1}{2} , -1 \\ 2^x = 2^{-1} , \xcancel{-1} \\ x = -1 \mathrm{อยู่ในช่วง} (-2, 0] \to \mathrm{ตอบ} 2 \end{gathered}$$