Matemática, perguntado por RowaBrasil, 9 meses atrás

resolva as equações do segundo grau abaixo, obtendo as respostas na forma de um número Complexo: a) x 2 + 2x + 10 = 0 b) x 2 + 6x + 25 = 0 c) x 2 – x + 2 = 0 d) x 2 + 25 = 0

Soluções para a tarefa

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Explicação passo-a-passo:

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

\sf \Delta=2^2-4\cdot1\cdot10

\sf \Delta=4-40

\sf \Delta=-36

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

\sf \Delta=36i^2

\sf x=\dfrac{-2\pm\sqrt{36i^2}}{2\cdot1}=\dfrac{-2\pm6i}{2}

\sf x'=\dfrac{-2+6i}{2}~\longrightarrow~x'=-1+3i

\sf x"=\dfrac{-2-6i}{2}~\longrightarrow~x"=-1-3i

\sf S=\{-1-3i,-1+3i\}

b) \sf x^2+6x+25=0

\sf \Delta=6^2-4\cdot1\cdot25

\sf \Delta=36-100

\sf \Delta=-64

\sf \Delta=64\cdot(-1)

\sf \Delta=64i^2

\sf x=\dfrac{-6\pm\sqrt{64i^2}}{2\cdot1}=\dfrac{-6\pm8i}{2}

\sf x'=\dfrac{-6+8i}{2}~\longrightarrow~x'=-3+4i

\sf x"=\dfrac{-6-8i}{2}~\longrightarrow~x"=-3-4i

\sf S=\{-3-4i,-3+4i\}

c) \sf x^2-x+2=0

\sf \Delta=(-1)^2-4\cdot1\cdot2

\sf \Delta=1-8

\sf \Delta=-7

\sf \Delta=7\cdot(-1)

\sf \Delta=7i^2

\sf x=\dfrac{-(-1)\pm\sqrt{7i^2}}{2\cdot1}=\dfrac{1\pm\sqrt{7}\cdot i}{2}

\sf x'=\dfrac{1+\sqrt{7}\cdot i}{2}

\sf x"=\dfrac{1-\sqrt{7}\cdot i}{2}

\sf S=\left\{\dfrac{1-\sqrt{7}\cdot i}{2},\dfrac{1+\sqrt{7}\cdot i}{2}\right\}

d) \sf x^2+25=0

\sf x^2=-25

\sf x^2=25\cdot(-1)

\sf x^2=25i^2

\sf x=\pm\sqrt{25i^2}

\sf x'=5i

\sf x"=-5i

\sf S=\{-5i,5i\}

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