Question

É sobre baskara alguém consegue me ajudar?

Answer 1

a)

$x^2-6x+10~=~0\\\\\\\Delta~=~(-6)^2-4.1.10~=~36-40~=~\boxed{-4}\\\\\\x'~=~\frac{6+\sqrt{-4}}{2~.~1}~=~\frac{6+\sqrt{-1~.~4}}{2}~=~\frac{6\,+\,\sqrt{-1}\,.\,\sqrt{4}}{2}~=~\frac{6\,+\,i\,.\,2}{2}~=~\frac{6}{2}+\frac{2i}{2}~=~\boxed{3+i}\\\\\\x''~=~\frac{6-\sqrt{-4}}{2~.~1}~=~\frac{6-\sqrt{-1~.~4}}{2}~=~\frac{6\,-\,\sqrt{-1}\,.\,\sqrt{4}}{2}~=~\frac{6\,-\,i\,.\,2}{2}~=~\frac{6}{2}-\frac{2i}{2}~=~\boxed{3-i}$

b)

$2x^2-4x+4~=~0\\\\\\\Delta~=~(-4)^2-4.2.4~=~16-32~=~\boxed{-16}\\\\\\x'~=~\frac{4+\sqrt{-16}}{2~.~2}~=~\frac{4+\sqrt{-1~.~16}}{4}~=~\frac{4\,+\,\sqrt{-1}\,.\,\sqrt{16}}{4}~=~\frac{4\,+\,i\,.\,4}{4}~=~\frac{4}{4}+\frac{4i}{4}~=~\boxed{1+i}\\\\\\x''~=~\frac{4-\sqrt{-16}}{2~.~2}~=~\frac{4-\sqrt{-1~.~16}}{4}~=~\frac{4\,-\,\sqrt{-1}\,.\,\sqrt{16}}{4}~=~\frac{4\,-\,i\,.\,4}{4}~=~\frac{4}{4}-\frac{4i}{4}~=~\boxed{1-i}$

c)

$x^2-6x+25~=~0\\\\\\\Delta~=~(-6)^2-4.1.25~=~36-100~=~\boxed{-64}\\\\\\x'~=~\frac{6+\sqrt{-64}}{2~.~1}~=~\frac{6+\sqrt{-1~.~64}}{2}~=~\frac{6\,+\,\sqrt{-1}\,.\,\sqrt{64}}{2}~=~\frac{6\,+\,i\,.\,8}{2}~=~\frac{6}{2}+\frac{8i}{2}~=~\boxed{3+4i}\\\\\\x''~=~\frac{6-\sqrt{-64}}{2~.~1}~=~\frac{6-\sqrt{-1~.~64}}{2}~=~\frac{6\,-\,\sqrt{-1}\,.\,\sqrt{64}}{2}~=~\frac{6\,-\,i\,.\,8}{2}~=~\frac{6}{2}-\frac{8i}{2}~=~\boxed{3-4i}$