Question

3x2-3=4×+2 como faz?

Answer 1

Olá!!!

Resolução!!!

3x² - 3 = 4x + 2

3x² - 3 - 4x - 2 = 0

3x² - 4x - 3 - 2 = 0

3x² - 4x - 5 = 0

a = 3, b = - 4, c= - 5

∆ = b² - 4ac
∆ = ( - 4 )² - 4 • 3 • ( - 5 )
∆ = 16 + 60
∆ = 76

x = - b ± √∆ / 2a
x = - ( - 4 ) ± √76 / 2 • 3

76 | 2
38 | 2
19 | 19
1

√2² • 19 = √76
2 • √19 = √76
2√19 = √76

x = 4 ± 2√19 / 6
x' = 4 + 2√19
x" = 4 - 2√19

S = { 4 - 2√19, 4 + 2√19 }

Espero ter ajudado!!,

Answer 2

3x² - 3 = 4x + 2
3x² - 4x - 3 - 2 = 0
3x² - 4x - 5 = 0

a = 3
b = -4
c = -5

Δ = b² -4 .a .c
Δ = (-4)² -4 . 3 . (-5)
Δ = 16  - 12 . (-5)
Δ = 16 + 60
Δ = 76

$\mathtt{\dfrac{-b+-\sqrt{\Delta}}{2~.~a}~~=~~\dfrac{-(-4)+-\sqrt{76}}{2~.~3}~~=~~\dfrac{4 +-\sqrt{76}}{6}}$

Fatora : 76

76     |2
38     |2
19     |19
1

2 . 2 . 19
2² . 19

$\mathtt{\dfrac{4 +-\sqrt{2^2~.~19}}{6}~~=~~\dfrac{4+-2\sqrt{19}}{6}} \\ \\ \\ \mathtt{x' = \dfrac{4 + 2\sqrt{19}}{6}~~(\div2)=~~\dfrac{\sqrt{19}+ 2}{3}} \\ \\ \\ \mathtt{x'' =\dfrac{4 - 2\sqrt{19}}{6}~~(\div2)=~~ -\dfrac{\sqrt{19}-2}{3}} \\ \\ \\ \\ \boxed{\boxed{\mathtt{Resposta: S:~~ -\dfrac{\sqrt{19}-2}{3}~~e~~\dfrac{\sqrt{19}+2}{3}}}}$