Question

Resolva as seguintes equações, utilizando a fórmula resolutiva da equação do 2º grau :

 

a) x² - x - 2 = 0 
x1 =
x2=

-----------------------------

b) x² + 9x + 8 = 0
x1=
x2=

--------------------------

c) x² - x - 20 = 0
x1=
x2=

-------------------------

d) x² - 8x + 7 = 0
x1=
x2=

--------------------------

e) x² - 3x - 4 = 0
x1=
x2=

--------------------------

com os calcúlos pra hoje por favor , URGEEEEEEEENTE GENTE  :/ 

Answer 1

a) $\text{x}^2-\text{x}-2=0$

 

$\text{x}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-2)}}{2\cdot1}=\dfrac{1\pm3}{2}$

 

$\text{x}_1=\dfrac{1+3}{2}=2$

 

$\text{x}_2=\dfrac{1-3}{2}=-1$

 

 

b) $\text{x}^2+9\text{x}+8=0$

 

$\text{x}=\dfrac{-9\pm\sqrt{9^2-4\cdot1\cdot8}}{2\cdot1}=\dfrac{-9\pm7}{2}$

 

$\text{x}_1=\dfrac{-9+7}{2}=-1$

 

$\text{x}_2=\dfrac{-9-7}{2}=-8$

 

 

c) $\text{x}^2-\text{x}-20=0$

 

$\text{x}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-20)}}{2\cdot1}=\dfrac{1\pm9}{2}$

 

$\text{x}_1=\dfrac{1+9}{2}=5$

 

$\text{x}_2=\dfrac{1-9}{2}=-4$

 

 

d) $\text{x}^2-8\text{x}+7=0$

 

$\text{x}=\dfrac{-(-8)\pm\sqrt{(-8)^2-4\cdot1\cdot7}}{2\cdot1}=\dfrac{8\pm6}{2}$

 

$\text{x}_1=\dfrac{8+6}{2}=7$

 

$\text{x}_2=\dfrac{8-6}{2}=1$

 

 

e) $\text{x}^2-3\text{x}-4=0$

 

$\text{x}=\dfrac{-(-3)\pm\sqrt{(-3)^2-4\cdot1\cdot(-4)}}{2\cdot1}=\dfrac{3\pm5}{2}$

 

$\text{x}_1=\dfrac{3+5}{2}=4$

 

$\text{x}_2=\dfrac{3-5}{2}=-1$