Respostas

  • Usuário do Brainly
2014-08-21T00:52:04-03:00
X² -x -3=0

a=1
b=-1
c=-3

Δ=b²-4ac
Δ=1²-(-3)
Δ=1+12
Δ=13
√Δ=± √13

x=(-b ±√Δ)/2a

x'= \frac{1+ \sqrt{13} }{2}

x"= \frac{1- \sqrt{13} }{2}

S={  \frac{1+ \sqrt{13} }{2} , \frac{1- \sqrt{13} }{2} }
  • Usuário do Brainly
2014-08-21T18:10:32-03:00
Temos que, x=\dfrac{-b\pm\sqrt{\Delta}}{2a}, onde:

\Delta=b^2-4\cdot a\cdot c

Queremos resolver a equação x^2-x-3=0.

Observe que, a=1, b=-1 e c=-3.

Assim, \Delta=(-1)^2-4\cdot1\cdot(-3)=1+12=13.

Deste modo, x=\dfrac{-(-1)\pm\sqrt{13}}{2\cdot1}=\dfrac{1\pm\sqrt{13}}{2}.

Logo, x'=\dfrac{1+\sqrt{13}}{2} e x"=\dfrac{1-\sqrt{13}}{2}.