Question

-3 X elevado ao quadrado menos 8x + 3 = 0​

Answer 1

Resposta:

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~-3x^{2}+8x+3=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=-3{;}~b=8~e~c=3\\\\\Delta=(b)^{2}-4(a)(c)=(8)^{2}-4(-3)(3)=64-(-36)=100\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(8)-\sqrt{100}}{2(-3)}=\frac{-8-10}{-6}=\frac{-18}{-6}=3\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(8)+\sqrt{100}}{2(-3)}=\frac{-8+10}{-6}=\frac{2\div2}{-6\div2}=-\frac{1}{3}\\\\S=\{3,~-\frac{1}{3}\}$

Answer 2

resolução!

$- 3 {x}^{2} - 8x + 3 = 0 \times ( - 1) \\ 3 {x}^{2} + 8x - 3 = 0$

∆ = 8^2 - 4 * 3 * (-3)

∆ = 64 + 36

∆ = 100

∆ =√100

∆ = 10

X ' = - 8 + 10/6

X ' = 2/6

X ' = 1/3

X " = - 8 - 10/6

X " = - 18/6

X " = - 3

S = { - 3 , 1/3 }