Question

Resolva as equações de 2° grau utilizando a fórmula de Bhaskara.

a) x²+x+2=0
b) x²-7x+10=0
c) x²+5x+6=0
d) x²+2x+4=0​

Answer 1

Explicação passo-a-passo:

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

$\sf \Delta=1^2-4\cdot1\cdot2$

$\sf \Delta=1-8$

$\sf \Delta=-7$

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

$\sf S=\{~\}$

b) $\sf x^2-7x+10=0$

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

$\sf \Delta=49-40$

$\sf \Delta=9$

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

$\sf x'=\dfrac{7+3}{2}~\rightarrow~x'=\dfrac{10}{2}~\rightarrow~x'=5$

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

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

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

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

$\sf \Delta=25-24$

$\sf \Delta=1$

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

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

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

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

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

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

$\sf \Delta=4-16$

$\sf \Delta=-12$

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

$\sf S=\{~\}$