Question

ME AJUDEM! "Resolva as equações a seguir usando a forma resolutiva:"
a) x² - 3x -28 = 0
b) x² + 12x + 36 = 0
c) 6x² – x – 1 = 0
d) x² - 4x - 4 =

Answer 1

para resolver esta tarefa basta aplicar bhaskara:

$\boxed{x = \dfrac{-b\pm \sqrt{b^2 - 4\cdot a\cdot c}}{2\cdot a}}$

A equação tem a seguinte fórmula : $\boxed{ax^2+bx + c =}$, agora basta substituir usando os dados das alternativas.

a) $x^2 -3x-28 = 0$

$x = \frac{-(-3)\pm \sqrt{-3^2 - 4\cdot 1\cdot -28}}{2\cdot 1}$

$x = \frac{3\pm \sqrt{9 + 112}}{2} = \frac{3\pm \sqrt{121}}{2} = \frac{3\pm 11}{2}$

$x_1 = \dfrac{3+11}{2} = \dfrac{14}{2}$  e $x_2 = \dfrac{3-11}{2} = \dfrac{-8}{2}$

R: $\boxed{x_1 = 7 \,\,\,\,x_2=-4}$

b) $x^2 + 12x + 36 = 0$

$x = \frac{-12\pm \sqrt{12^2 - 4\cdot 1\cdot 36}}{2\cdot 1}$

$x = \frac{-12\pm \sqrt{144 - 144}}{2} = \frac{-12\pm \sqrt{0}}{2} = \frac{-12}{2}$

R: $\boxed{x = -6}$

c) $6x^2 - x - 1 = 0$

$x = \frac{-(-1)\pm \sqrt{-1^2 - 4\cdot 6\cdot -1}}{2\cdot 6}$

$x = \frac{1\pm \sqrt{1 +24}}{12} = \frac{1\pm \sqrt{25}}{12} = \frac{1\pm 5}{12}$

$x_1 = \frac{1+5}{12} = \frac{6}{12} \,\,\,\, x_2 = \frac{1-5}{12} = \frac{-4}{12}$

R: $\boxed{x_1 = \frac{1}{2} \,\,\,\,\, x_2= -\frac{1}{3}}$

d) $x^2 - 4x - 4 = 0$

$x = \frac{-(-4)\pm \sqrt{-4^2 - 4\cdot 1\cdot -4}}{2\cdot 1}$

$x = \frac{4\pm \sqrt{16 + 16}}{2} = \frac{4\pm \sqrt{32}}{2}$

R: $\boxed{x_1 = \frac{4+\sqrt{32}}{2} \,\,\,\,\, x_2 = \frac{4-\sqrt{32}}{2} }$