Jika α dan β adalah solusi persamaan eksponen [tex] 2^{x+1} - 3. 2^{ \frac{1}{2}x } +1=0[/tex] , maka αβ = A. 0 B. 1 C. 2 D. 3 E. 4 tolong jawab pakai jalan ya

[tex] {2}^{x + 1} - 3. {2}^{ \frac{1}{2}x } + 1 = 0 \\ 2. {2}^{x} - 3. {2}^{ \frac{1}{2}x } + 1 = 0 \\ \\ misal \: {2}^{ \frac{1}{2} x} = a \\ \\ 2 {a}^{2} - 3a + 1 = 0 \\ (2a - 1)(a - 1) = 0 \\ \\ 2a - 1 = 0 \\ a = \frac{1}{2} = {2}^{ - 1} \\ \\ {2}^{ \frac{1}{2}x } = {2}^{ - 1} \\ \frac{1}{2} x = - 1 \\ \\ x = - 2 \\atau \\ a - 1 = 0 \\ a = 1 = {2}^{0} \\ {2}^{ \frac{1}{2} x} = {2}^{0} \\ x = 0 \\ \\ \alpha = 2 \\ \beta = 0 \\ \\ \alpha \beta = 2 \times 0 \\ \alpha \beta = 0[/tex]
Jawaban A