Question

Resolva as funções: A) x elevado a 2 - 5x + 6 B) x elevado a 2 - 8 x + 12 C) x elevado a 2 + 2x - 8 D) x elevado a 2 - 5 × + 8 E) 2 x elevado a 2 - 8 x + 8 F) x elevado a 2 - 4x - 5

Answer 1

$a) \\ x^2-5x+6=0 \\ x_{12} = \frac{-(-5)+/- \sqrt{(-5)^2-4(1)(6)} }{2(1)} = \frac{5+/- \sqrt{25-24} }{2} =\frac{5+/- 1 }{2} \\ x_{1} = \frac{6}{2} =3 \\ x_{2} = \frac{4}{2} =2 \\ b) \\ x^2-8x+12=0 \\ x_{12} = \frac{-(-8)+/- \sqrt{(-8)^2-4(1)(12)} }{2(1)} = \frac{8+/- \sqrt{64-48} }{2} =\frac{8+/- \sqrt{16} }{2} =\frac{8+/- 4 }{2} \\ x_{1} = \frac{12}{2} =6 \\ x_{2} = \frac{4}{2} =2$

$c) \\ x^2+2x-8=0 \\ x_{12} = \frac{-(2)+/- \sqrt{(2)^2-4(1)(-8)} }{2(1)} = \frac{-2+/- \sqrt{4+32} }{2} =\frac{-2+/- \sqrt{36} }{2} =\frac{-2+/- 6 }{2} \\ x_{1} = \frac{-2+6}{2} =2 \\ x_{2} = \frac{-2-6}{2} =-4 \\ \\ d) \\ x^2-5x+8=0 \\ x_{12} = \frac{-(-5)+/- \sqrt{(-5)^2-4(1)(8)} }{2(1)} = \frac{5+/- \sqrt{25-32} }{2} =\frac{5+/- \sqrt{-7} }{2} ==\frac{5+/- i*\sqrt{7} }{2}\\ x_{1} = \frac{5+i*\sqrt{7}}{2} \\ x_{2} = \frac{5-i*\sqrt{7}}{2} \\ onde: \\ i= \sqrt{-1}$

$e) \\ 2x^2-8x+8=0 \\ x^2-4x+4=0 \\x_{12} = \frac{-(-4)+/- \sqrt{(-4)^2-4(1)(4)} }{2(1)} = \frac{4+/- 0 }{2} =\frac{4}{2} \\ x_{1} = x_{2} =2$

$f) \\ x^2-4x-5=0 \\ x_{12} = \frac{-(-4)+/- \sqrt{(-4)^2-4(1)(-5)} }{2(1)} = \frac{4+/- \sqrt{16+20} }{2} =\frac{4+/- \sqrt{36} }{2} =\frac{4+/- 6 }{2} \\ x_{1} = \frac{4+6}{2} =5 \\ x_{2} = \frac{4-6}{2} =-1$