Question

Resolva as equações do 2° grau abaixo:

a) x² - 5 + 6 = 0

b) x² - 8x + 12 = 0

c) x² + 2x - 8 = 0

d) x² - 5x + 8 = 0

e) 2x² - 8x + 8 = 0

f) x² - 4x - 5 = 0

g) -x² + x + 12 = 0

h) -x² + 6x - 5 = 0

i) 6x² + x - 1 = 0

J) 3x² - 7x + 2 = 0

me ajudem, eu imploro por favor



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Answer 1

Explicação passo-a-passo:

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

$\sf \Delta=(-5)^2-4\cdot1\cdot6$

$\sf \Delta=25-24$

$\sf \Delta=1$

$\sf x=\dfrac{-(-5)\pm\sqrt{1}}{2\cdot1}=\dfrac{5\pm1}{2}$

$\sf x'=\dfrac{5+1}{2}~\longrightarrow~x'=\dfrac{6}{2}~\longrightarrow~x'=3$

$\sf x"=\dfrac{5-1}{2}~\longrightarrow~x"=\dfrac{4}{2}~\longrightarrow~x"=2$

$\sf S=\{2,3\}$

b) $\sf x^2-8x+12=0$

$\sf \Delta=(-8)^2-4\cdot1\cdot12$

$\sf \Delta=64-48$

$\sf \Delta=16$

$\sf x=\dfrac{-(-8)\pm\sqrt{16}}{2\cdot1}=\dfrac{8\pm4}{2}$

$\sf x'=\dfrac{8+4}{2}~\longrightarrow~x'=\dfrac{12}{2}~\longrightarrow~x'=6$

$\sf x"=\dfrac{8-4}{2}~\longrightarrow~x"=\dfrac{4}{2}~\longrightarrow~x"=2$

$\sf S=\{2,6\}$

c) $\sf x^2+2x-8=0$

$\sf \Delta=2^2-4\cdot1\cdot(-8)$

$\sf \Delta=4+32$

$\sf \Delta=36$

$\sf x=\dfrac{-2\pm\sqrt{36}}{2\cdot1}=\dfrac{-2\pm6}{2}$

$\sf x'=\dfrac{-2+6}{2}~\longrightarrow~x'=\dfrac{4}{2}~\longrightarrow~x'=2$

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

$\sf S=\{-4,2\}$

d) $\sf x^2-5x+8=0$

$\sf \Delta=(-5)^2-4\cdot1\cdot8$

$\sf \Delta=25-32$

$\sf \Delta=-7$

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$\sf S=\{~~\}$

e) $\sf 2x^2-8x+8=0$

$\sf x^2-4x+4=0$

$\sf \Delta=(-4)^2-4\cdot1\cdot4$

$\sf \Delta=16-16$

$\sf \Delta=0$

$\sf x=\dfrac{-(-4)\pm\sqrt{0}}{2\cdot1}=\dfrac{4\pm0}{2}$

$\sf x'=x"=\dfrac{4}{2}$

$\sf x'=x"=2$

$\sf S=\{2\}$

f) $\sf x^2-4x-5=0$

$\sf \Delta=(-4)^2-4\cdot1\cdot(-5)$

$\sf \Delta=16+20$

$\sf \Delta=36$

$\sf x=\dfrac{-(-4)\pm\sqrt{36}}{2\cdot1}=\dfrac{4\pm6}{2}$

$\sf x'=\dfrac{4+6}{2}~\longrightarrow~x'=\dfrac{10}{2}~\longrightarrow~x'=5$

$\sf x"=\dfrac{4-6}{2}~\longrightarrow~x"=\dfrac{-2}{2}~\longrightarrow~x"=-1$

$\sf S=\{-1,5\}$

g) $\sf -x^2+x+12=0$

$\sf \Delta=1^2-4\cdot(-1)\cdot12$

$\sf \Delta=1+48$

$\sf \Delta=49$

$\sf x=\dfrac{-1\pm\sqrt{49}}{2\cdot(-1)}=\dfrac{-1\pm7}{-2}$

$\sf x'=\dfrac{-1+7}{-2}~\longrightarrow~x'=\dfrac{6}{-2}~\longrightarrow~x'=-3$

$\sf x"=\dfrac{-1-7}{-2}~\longrightarrow~x"=\dfrac{-8}{-2}~\longrightarrow~x"=4$

$\sf S=\{-3,4\}$

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

$\sf \Delta=6^2-4\cdot(-1)\cdot(-5)$

$\sf \Delta=36-20$

$\sf \Delta=16$

$\sf x=\dfrac{-6\pm\sqrt{16}}{2\cdot(-1)}=\dfrac{-6\pm4}{-2}$

$\sf x'=\dfrac{-6+4}{-2}~\longrightarrow~x'=\dfrac{-2}{-2}~\longrightarrow~x'=1$

$\sf x"=\dfrac{-6-4}{-2}~\longrightarrow~x"=\dfrac{-10}{-2}~\longrightarrow~x"=5$

$\sf S=\{1,5\}$

i) $\sf 6x^2+x-1=0$

$\sf \Delta=1^2-4\cdot6\cdot(-1)$

$\sf \Delta=1+24$

$\sf \Delta=25$

$\sf x=\dfrac{-1\pm\sqrt{25}}{2\cdot6}=\dfrac{-1\pm5}{12}$

$\sf x'=\dfrac{-1+5}{12}~\longrightarrow~x'=\dfrac{4}{12}~\longrightarrow~x'=\dfrac{1}{3}$

$\sf x"=\dfrac{-1-5}{12}~\longrightarrow~x"=\dfrac{-6}{12}~\longrightarrow~x"=\dfrac{-1}{2}$

$\sf S=\left\{\dfrac{-1}{2},\dfrac{1}{3}\right\}$

j) $\sf 3x^2-7x+2=0$

$\sf \Delta=(-7)^2-4\cdot3\cdot2$

$\sf \Delta=49-24$

$\sf \Delta=25$

$\sf x=\dfrac{-(-7)\pm\sqrt{25}}{2\cdot3}=\dfrac{7\pm5}{6}$

$\sf x'=\dfrac{7+5}{6}~\longrightarrow~x'=\dfrac{12}{6}~\longrightarrow~x'=2$

$\sf x"=\dfrac{7-5}{6}~\longrightarrow~x"=\dfrac{2}{6}~\longrightarrow~x"=\dfrac{1}{3}$

$\sf S=\left\{\dfrac{1}{3},2\right\}$