Question

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Answer 1

Ответ:

${x}^{2} + {3}^{ log_{3}(x) } = 12$

$x \leqslant 0$

${x}^{2} + {3}^{ log_{3}(x) } = 12.x > 0$

${x}^{2} + x = 12$

$x \times (x + 1) = 12$

${x}^{2} + x = 12$

${x}^{2} + x - 12 = 0$

${x}^{2} + 4x - 3x - 12 = 0$

$x \times (x + 4) - 3(x + 4) = 0$

$(x + 4) \times (x - 3) = 0$

$x + 4 = 0 \\ x - 3 = 0$

$x + 4 = 0 \\ x + 4 - 4 = 0 - 4 \\ x = - 4$

$x - 3 = 0 \\ x - 3 + 3 = 0 + 3 \\ x = 3$

$x > 0$

$x = - 4 \\ x = 3 \\ x > 0$

$x = 3$