Question

resolva as equação do segundo grau incompletas e completas alguém me ajuda?​

Answer 1

Explicação passo-a-passo:

1)

a)

$\sf x^2-10x=0$

$\sf x\cdot(x-10)=0$

• $\sf x'=0$

• $\sf x-10=0~\rightarrow~x"=10$

$\sf S=\{0,10\}$

b)

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

$\sf x\cdot(x+9)=0$

• $\sf x'=0$

• $\sf x+9=0~\rightarrow~x"=-9$

$\sf S=\{-9,0\}$

c)

$\sf x^2-625=0$

$\sf x^2=625$

$\sf x=\pm\sqrt{625}$

• $\sf x'=25$

• $\sf x"=-25$

$\sf S=\{-25,25\}$

d)

$\sf y^2-400=0$

$\sf y^2=400$

$\sf y=\pm\sqrt{400}$

• $\sf y'=20$

• $\sf y"=-20$

$\sf S=\{-20,20\}$

2)

a)

$\sf x^2+3x-10=0$

$\sf \Delta=3^2-4\cdot1\cdot(-10)$

$\sf \Delta=9+40$

$\sf \Delta=49$

$\sf x=\dfrac{-3\pm\sqrt{49}}{2\cdot1}=\dfrac{-3\pm7}{2}$

$\sf x'=\dfrac{-3+7}{2}~\rightarrow~x'=\dfrac{4}{2}~\rightarrow~x'=2$

$\sf x"=\dfrac{-3-7}{2}~\rightarrow~x"=\dfrac{-10}{2}~\rightarrow~x"=-5$

$\sf S=\{-5,2\}$

b)

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

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

$\sf \Delta=1+24$

$\sf \Delta=25$

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

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

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

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

c)

$\sf 2x^2-4x+3=0$

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

$\sf \Delta=16-24$

$\sf \Delta=-8$

Não há raízes reais

$\sf S=\{~\}$