Question

04) Resolver em Ras equações do 2º grau utilizando o discriminante (delta A) encontrado no
exercicio anterior dando a solução.
EQUAÇÕES COMPLETAS DO TIPO: a²+bx+c = 0, com (a, b e c diferente de 0). COM CÁLCULOS POR FAVOR

EX: x²-6x+9=0
- S = {3}

a) x² + 6x + 5 = 0
S= ?

b) 2x² - 7x + 3 = 0
S= ?

c) x² - 4x + 4 = 0
S= ?

d) 3x² - 2x + 4 = 0
S= ?

e)-3x² - 2x + 1 = 0
S= ?​​

Answer 1

Explicação passo-a-passo:

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

$\sf \Delta=6^2-4\cdot1\cdot5$

$\sf \Delta=36-20$

$\sf \Delta=16$

$\sf x=\dfrac{-6\pm\sqrt{16}}{2\cdot1}=\dfrac{-6\pm4}{2}$

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

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

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

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

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

$\sf \Delta=49-24$

$\sf \Delta=25$

$\sf x=\dfrac{-(-7)\pm\sqrt{25}}{2\cdot2}=\dfrac{7\pm5}{4}$

$\sf x'=\dfrac{7+5}{4}~\rightarrow~x'=\dfrac{12}{4}~\rightarrow~x'=3$

$\sf x"=\dfrac{7-5}{4}~\rightarrow~x"=\dfrac{2}{4}~\rightarrow~x'=\dfrac{1}{2}$

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

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

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

$\sf \Delta=16-16$

$\sf \Delta=0$

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

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

$\sf S=\{2\}$

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

$\sf \Delta=(-2)^2-4\cdot3\cdot4$

$\sf \Delta=4-48$

$\sf \Delta=-44$

Como $\sf \Delta < 0$, não há raízes reais

$\sf S=\{~\}$

e) $\sf -3x^2-2x+1=0$

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

$\sf \Delta=4+12$

$\sf \Delta=16$

$\sf x=\dfrac{-(-2)\pm\sqrt{16}}{2\cdot(-3)}=\dfrac{2\pm4}{-6}$

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

$\sf x"=\dfrac{2-4}{-6}~\rightarrow~x"=\dfrac{-2}{-6}~\rightarrow~x'=\dfrac{1}{3}$

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