Question

Calcule as Raízes Imaginárias das Equações de Segundo Grau ( Equação Polinomial ) Conforme Material de Estudo Sobre Números Complexos

Answer 1

Explicação passo-a-passo:

1) $\sf x^2+9=0$

$\sf x^2=-9$

$\sf x^2=9\cdot(-1)$

$\sf x^2=9i^2$

$\sf x=\pm\sqrt{9i^2}$

$\sf x=\pm3i$

$\sf S=\{-3i,3i\}$

2) $\sf 2x^2+10=0$

$\sf 2x^2=-10$

$\sf x^2=\dfrac{-10}{2}$

$\sf x^2=-5$

$\sf x^2=5\cdot(-1)$

$\sf x^2=5i^2$

$\sf x=\pm\sqrt{5i^2}$

$\sf x=\pm\sqrt{5}\cdot i$

$\sf S=\{-\sqrt{5}\cdot i,\sqrt{5}\cdot i\}$

3) $\sf 2x^2-6x+9=0$

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

$\sf \Delta=(-6)^2-4\cdot2\cdot9$

$\sf \Delta=36-72$

$\sf \Delta=-36$

$\sf \Delta=36\cdot(-1)$

$\sf \Delta=36i^2$

$\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2\cdot a}$

$\sf y=\dfrac{-(-6)\pm\sqrt{36i^2}}{2\cdot2}=\dfrac{6\pm6i}{4}$

$\sf y'=\dfrac{6+6i}{4}~\longrightarrow~y'=\dfrac{3+3i}{2}$

$\sf y"=\dfrac{6-6i}{4}~\longrightarrow~y"=\dfrac{3-3i}{2}$

$\sf S=\left\{\dfrac{3-3i}{2},\dfrac{3+3i}{2}\right\}$

4) $\sf x^2-10x+34=0$

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

$\sf \Delta=(-10)^2-4\cdot1\cdot34$

$\sf \Delta=100-136$

$\sf \Delta=--36$

$\sf \Delta=36\cdot(-1)$

$\sf \Delta=36i^2$

$\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2\cdot a}$

$\sf y=\dfrac{-(-10)\pm\sqrt{36i^2}}{2\cdot1}=\dfrac{10\pm6i}{2}$

$\sf y'=\dfrac{10+6i}{4}~\longrightarrow~y'=5+3i$

$\sf y"=\dfrac{10-6i}{4}~\longrightarrow~y"=5-3i$

$\sf S=\{5-3i,5+3i\}$

5) $\sf 6x^2+216=0$

$\sf 6x^2=-216$

$\sf x^2=\dfrac{-216}{6}$

$\sf x^2=-36$

$\sf x^2=36\cdot(-1)$

$\sf x^2=36i^2$

$\sf x=\pm\sqrt{36i^2}$

$\sf x=\pm6i$

$\sf S=\{-6i,6i\}$