Question

como resolver essas equações do 2° grau completa na fórmula de BhasKara:

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​

Answer 1

Resposta:

Explicação passo-a-passo:

Well, Well, Well.........

Sabemos que

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

b)

$\frac{-\left(-8\right)+\sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:12}}{2\cdot \:1}$

$\frac{-\left(-8\right)-\sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:12}}{2\cdot \:1}$

X' = 6 e X' = 2

c)

$\frac{-2+\sqrt{2^2-4\cdot \:1\cdot \left(-8\right)}}{2\cdot \:1}</p><p>\frac{-2-\sqrt{2^2-4\cdot \:1\cdot \left(-8\right)}}{2\cdot \:1}$

X' = 2 e X' = -4

d)

Δ = b² - 4.a.c

Δ = -5² - 4 . 1 . 8

Δ = 25 - 4. 1 . 8

Δ = -7

Não há raízes reais.

e)

$\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:2\cdot \:8}}{2\cdot \:2}$

X' = 2 e X' = 2

f)

$\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:1\cdot \left(-5\right)}}{2\cdot \:1}$

$\frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:1\cdot \left(-5\right)}}{2\cdot \:1}$

X' = 5 e X' = -1

g)

$\frac{-1+\sqrt{1^2-4\left(-1\right)\cdot \:12}}{2\left(-1\right)}$

$\frac{-1-\sqrt{1^2-4\left(-1\right)\cdot \:12}}{2\left(-1\right)}$

X' = -3 e X' = 4

h)

$\frac{-6+\sqrt{6^2-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$

$\frac{-6-\sqrt{6^2-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$

X' = 1 e X' = 5

i)

$\frac{-1+\sqrt{1^2-4\cdot \:6\left(-1\right)}}{2\cdot \:6}$

$\frac{-1-\sqrt{1^2-4\cdot \:6\left(-1\right)}}{2\cdot \:6}$

X' = 1/3 e X' = -1/2

j)

$\frac{-\left(-7\right)+\sqrt{\left(-7\right)^2-4\cdot \:3\cdot \:2}}{2\cdot \:3}$

$\frac{-\left(-7\right)-\sqrt{\left(-7\right)^2-4\cdot \:3\cdot \:2}}{2\cdot \:3}$

X' = 2 e X' = 1/3