Matemática, perguntado por add87, 11 meses atrás

Resolvam essas equações de 2 grau pleaseee

Anexos:

Soluções para a tarefa

Respondido por Usuário anônimo
0

Explicação passo-a-passo:

a) z^2-2z+1=0

\Delta=(-2)^2-4\cdot1\cdot1

\Delta=4-4

\Delta=0

z=\dfrac{-(-2)\pm\sqrt{0}}{2\cdot1}=\dfrac{2\pm0}{2}

z'=z"=\dfrac{2}{2}

z'=z"=1

b) x^2-7x+10=0

\Delta=(-7)^2-4\cdot1\cdot10

\Delta=49-40

\Delta=9

x=\dfrac{-(-7)\pm\sqrt{9}}{2\cdot1}=\dfrac{7\pm3}{2}

x'=\dfrac{7+3}{2}~\longrightarrow~x'=\dfrac{10}{2}~\longrightarrow~x'=5

x"=\dfrac{7-3}{2}~\longrightarrow~x"=\dfrac{4}{2}~\longrightarrow~x"=2

c) x^2+23x+22=0

\Delta=23^2-4\cdot1\cdot22

\Delta=529-88

\Delta=441

x=\dfrac{-23\pm\sqrt{441}}{2\cdot1}=\dfrac{-23\pm21}{2}

x'=\dfrac{-23+21}{2}~\longrightarrow~x'=\dfrac{-2}{2}~\longrightarrow~x'=-1

x"=\dfrac{-23-21}{2}~\longrightarrow~x"=\dfrac{-44}{2}~\longrightarrow~x"=-22

d) x^2-3x-130=0

\Delta=(-3)^2-4\cdot1\cdot130

\Delta=9+520

\Delta=529

x=\dfrac{-(-3))\pm\sqrt{529}}{2\cdot1}=\dfrac{3\pm23}{2}

x'=\dfrac{3+23}{2}~\longrightarrow~x'=\dfrac{26}{2}~\longrightarrow~x'=13

x"=\dfrac{3-23}{2}~\longrightarrow~x"=\dfrac{-20}{2}~\longrightarrow~x"=-10

e) y^2+11y+18=0

\Delta=11^2-4\cdot1\cdot18

\Delta=121-72

\Delta=49

y=\dfrac{-11\pm\sqrt{49}}{2\cdot1}=\dfrac{-11\pm7}{2}

y'=\dfrac{-11+7}{2}~\longrightarrow~y'=\dfrac{-4}{2}~\longrightarrow~y"=-2

y"=\dfrac{-11-7}{2}~\longrightarrow~y"=\dfrac{-18}{2}~\longrightarrow~y"=-9

f) x^2-9x+20=0

\Delta=(-9)^2-4\cdot1\cdot20

\Delta=81-80

\Delta=1

x=\dfrac{-(-9)\pm\sqrt{1}}{2\cdot1}=\dfrac{9\pm1}{2}

x'=\dfrac{9+1}{2}~\longrightarrow~x'=\dfrac{10}{2}~\longrightarrow~x'=5

x"=\dfrac{9-1}{2}~\longrightarrow~x"=\dfrac{8}{2}~\longrightarrow~x"=4

g) x^2+3x+36=0

\Delta=3^2-4\cdot1\cdot36

\Delta=9-144

\Delta=-135

Não há raízes reais

h) x^2-4x-21=0

\Delta=(-4)^2-4\cdot1\cdot(-21)

\Delta=16+84

\Delta=100

x=\dfrac{-(-4)\pm\sqrt{100}}{2\cdot1}=\dfrac{4\pm10}{2}

x'=\dfrac{4+10}{2}~\longrightarrow~x'=\dfrac{14}{2}~\longrightarrow~x'=7

x"=\dfrac{4-10}{2}~\longrightarrow~x"=\dfrac{-6}{2}~\longrightarrow~x"=-3

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