Question

Rezolvati in multimea numerelor reale ecuata 2^x^2+4x+2=64 x 2^x.

Answer 1

Răspuns:

Explicație pas cu pas:

$2 {}^{x {}^{2} + 4x + 2 } = 64 \times 2 {}^{x}$

$2 {}^{x {}^{2} + 4x + 2 } = 2 {}^{6} \times 2 {}^{x}$

$2 {}^{x {}^{2} + 4x + 2 } = 2 {}^{x + 6}$

$x {}^{2} + 4x + 2 = x + 6$

$x {}^{2} + 4x + 2 - x - 6 = 0$

$x {}^{2} + 3x - 4 = 0$

$a = 1 \: , \: b = 3 \: , \: c = - 4$

$\Delta = b {}^{2} - 4ac = 3 {}^{2} - 4 \times 1 \times ( - 4) \\ = 9 - ( - 16) = 9 +16 = 25 > 0 \Longrightarrow \exists \: \: x_1 \: , \: x_2 \in R \: \: ; \: \: x_1 \neq x_2$

$x_1 = \frac{ - b - \sqrt{ \Delta } }{2a} = \frac{ - 3 - \sqrt{25} }{2} = \frac{ - 3 - 5}{2} = \frac{ - 8}{2} = - 4$

$x_2 = \frac{ - b + \sqrt{\Delta} }{2a} = \frac{ - 3 + \sqrt{25} }{2} = \frac{ - 3 + 5}{2} = \frac{2}{2} = 1$

${S = \Big\{ - 4 \: , \: 1 \Big\}}$