Question

Resolva em Bhaskara ou Delta

Por favor galera preciso muito resposta passo a passo.. me ajudem..

Answer 1

Resposta:

Explicação passo-a-passo:

7) a=-1    b=1   c=12

$x=\dfrac{-b\pm\sqrt{b^{2}-4.a.c}}{2.a}\\\\\\x=\dfrac{-1\pm\sqrt{1^{2}-4.(-1).12}}{2.(-1)}\\\\\\x=\dfrac{-1\pm\sqrt{1+48}}{-2}\\\\\\x=\dfrac{-1\pm\sqrt{49}}{-2}\\\\\\x=\dfrac{-1\pm7}{-2}\\\\\\x'=\dfrac{-1+7}{-2}=\dfrac{6}{-2}=-3\\\\\\x"=\dfrac{-1-7}{-2}=\frac{-8}{-2}=4$

8) a=-1   b=6    c=-5

$x=\dfrac{-b\pm\sqrt{b^{2}-4.a.c}}{2.a}\\\\\\x=\dfrac{-6\pm\sqrt{6^{2}-4.(-1).(-5)}}{2.(-1)}\\\\\\x=\dfrac{-6\pm\sqrt{36-20}}{-2}\\\\\\x=\dfrac{-6\pm\sqrt{16}}{-2}\\\\\\x=\dfrac{-6\pm4}{-2}\\\\\\x'=\dfrac{-6+4}{-2}=\dfrac{-2}{-2}=1\\\\\\x"=\dfrac{-6-4}{-2}=\dfrac{-10}{-2}=5$

9) a = 6    b = 1    c = -1

$x=\dfrac{-b\pm\sqrt{b^{2}-4.a.c}}{2.a}\\\\\\x=\dfrac{-1\pm\sqrt{1^{2}-4.6.(-1)}}{2.6}\\\\\\x=\dfrac{-1\pm\sqrt{1+24}}{12}\\\\\\x=\dfrac{-1\pm\sqrt{25}}{12}\\\\\\x=\dfrac{-1\pm5}{12}\\\\\\x'=\dfrac{-1+5}{12}=\dfrac{4}{12}=\dfrac{1}{3}\\\\\\x"=\dfrac{-1-5}{12}=\dfrac{-6}{12}=-\dfrac{1}{2}$

10)  a=3    b = -7   c=2

$x=\dfrac{-b\pm\sqrt{b^{2}-4.a.c}}{2.a}\\\\\\x=\dfrac{-(-7)\pm\sqrt{(-7)^{2}-4.3.2}}{2.3}\\\\\\x=\dfrac{7\pm\sqrt{49-24}}{6}\\\\\\x=\dfrac{7\pm\sqrt{25}}{6}\\\\\\x=\dfrac{7\pm5}{6}\\\\\\x'=\dfrac{7+5}{6}=\dfrac{12}{6}=2\\\\\\\\x"=\dfrac{7-5}{6}=\dfrac{2}{6}=\dfrac{1}{3}$