Question

pf me ajudem nessa conta de matemática

Answer 1

$x^{4}+6x^{3}+9x^{2}-4x^{2}-12x+3 = 0 \\ (x^{2})^{2}+2.x^{2}.3x+(3x)^{2} -4.(x^{2}+3x)+3 = 0 \\ (x^{2}+3x)^{2} -4.(x^{2}+3x)+3 = 0 \\ (x^{2} + 3x) = y \\ y^{2} - 4y + 3 = 0 \\ y' = 3 \\ y" = 1 \\ x^{2} + 3x = 3 \\ x^{2} + 3x - 3 = 0 \\ x' = \frac{(-3+ \sqrt{21})}{2} \\ x" = \frac{(-3- \sqrt{21})}{2} \\ ou \\ x^{2} + 3x = 1 \\ x^{2} + 3x - 1 = 0 \\ x' = \frac{(-3+ \sqrt{13})}{2} \\ x" = \frac{(-3- \sqrt{13})}{2}$

Resposta: $</span><span>(\frac{(-3+ \sqrt{21})}{2}; \frac{(-3- \sqrt{21})}{2}; \frac{(-3+ \sqrt{13})}{2}; \frac{(-3- \sqrt{13})}{2})$