Question

1) RESOLVA AS EQUAÇÕES DE 2º GRAU
a) 3x2 - 7x + 2 = 0

b) 2x2 - 7x = 15
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Answer 1

Resposta:

a) 3x^2 - 7x + 2 = 0

Δ = 49 - 4 (3) (2)

Δ = 49 - 24

Δ = 25

x1 = (7 + 5) / 6

x1 = 12/6 = 2

x2 = (7 - 5) / 6

x2 = 2/6 = 1/3

S = {2, 1/3}

b) 2x^2 - 7x - 15 = 0

Δ = 49 - 4 (2) (-15)

Δ = 169

x1 = (7 + 13) / 4

x1 = 20/4 = 5

x2 = (7 - 13) / 4

x2 = -6/4 = -3/2

S = {5, -3/2}

Quero melhor resposta ^^

Answer 2

Em ambas, vamos utilizar a fórmula quadrática (Bhaskara):

$\boldsymbol{\boxed{x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}}}$

a) 3x² - 7x + 2 = 0,  com a = 3,   b = -7   e  c = 2

$\boldsymbol{x=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\times 3\times 2}}{2\times 3}\to}\\\\\\\boldsymbol{x=\dfrac{7\pm\sqrt{49-24}}{6}\to x= \dfrac{7\pm 5}{6}\to}\\\\\\\boldsymbol{x_{1}=\dfrac{7+5}{6}\to\boxed{x_{1}=2}\,\,\checkmark}\\\\\\\boldsymbol{x_{2}=\dfrac{7-5}{6}\to\boxed{x_{2}=\dfrac{1}{3}}\checkmark}$

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b) 2x² - 7x - 15 = 0,  com  a = 2,   b = -7   e c = -15

$\boldsymbol{x=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\times 2\times (-15)}}{2\times 2}\to}\\\\\\\boldsymbol{x=\dfrac{7\pm\sqrt{49+120}}{4}\to x= \dfrac{7\pm 13}{4}\to}\\\\\\\boldsymbol{x_{1}=\dfrac{7+13}{4}\to\boxed{x_{1}=5}\,\,\checkmark}\\\\\\\boldsymbol{x_{2}=\dfrac{7-13}{4}\to\boxed{x_{2}=-\dfrac{3}{2}}\checkmark}$

É isso!! :(