Valendo 100 pontos!resolva as equações do 2° grau a) x²+8x+15=0b) x (x+2)=1c) (x+2)²=4.... Pergunta de ideia deMoisesjunio

Comentários Formulas - Δ = b²- 4 . a . c / x = - b+ - √ / 2 .a

a)x²+8x+15=0

a = x
b = 8x
c = 15

Δ = b²- 4 . a . c
Δ = (8)² - 4 . 1 . 15
Δ = 64 - 60
Δ = 4

x = - b+ - √ / 2 .a

x = -8 +- √4
-----------------
2 . 1

x = - 8 + - 2
---------------
2

I= - 8 + 2 = -6 / 2 = - 3
II = - 8 - 2 = -10/2 = - 5

s = (-3,-5)

b) x (x+2)=1
x² + 2x = 1
x² + 2x - 1 = 0

4 + 4
Δ = 8

c)(x+2)²=4

x² + 2.x.2 + 2²= 4
x² + 4x + 4 = 4
x² + 4x = 0

x = 0
x = - 4 = 0
x = 4

s = ( 0 , 4 )

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Comentários Olá Moisés, tudo bem?

Utilize a fórmula de Bhaskara par resolver:
$x=\frac{-b\pm \sqrt{b^{2}-4\times a \times c} }{2\times a}$

Nas alternativas B e C faça a resolução das equações antes de resolver com Bhaskara.

Vamos ao cálculo

A) A= 1, B= 8; C= 15.
$x=\frac{-b\pm \sqrt{b^{2}-4\times a \times c} }{2\times a}\\\\ x=\frac{-8\pm \sqrt{8^{2}-4\times 1 \times 15} }{2\times 1}\\\\ x=\frac{-8\pm \sqrt{64-60} }{2}\\\\ x=\frac{-8\pm \sqrt{4} }{2}\\\\ x=\frac{-8\pm 2 }{2}\\\\ x'= \frac{-8+2}{2}=\frac{-6}{2}= \boxed{-3} \\\\ x''=\frac{-8-2}{2}= \frac{-10}{2}= \boxed{-5}\\\\ S=\{-5\:,\:-3\}$

S={-5, -3}

B)
Resolvendo a equação
$x \times (x+2)=1\\
x^{2}+2x=1\\
x^{2}+2x-1=0$

A= 1; B= 2, C= -1
$x=\frac{-b\pm \sqrt{b^{2}-4\times a \times c} }{2\times a}\\\\
x=\frac{-2\pm \sqrt{2^{2}-4\times 1 \times (-1)} }{2\times 1}\\\\
x=\frac{-2\pm \sqrt{4+4} }{2}\\\\ x=\frac{-8\pm \sqrt{8} }{2}\\\\
x=\frac{-8\pm 2\sqrt{2} }{2}\\\\
x'= \frac{-2+2\sqrt{2}}{2}= \boxed{-1+\sqrt{2} } \\\\
x''=\frac{-8-2\sqrt{2}}{2}= \boxed{-1-\sqrt{2} }\\\\
S=\{-1 - \sqrt{2}\:,\:-1+ \sqrt{2}\}$

S={ -1-√2, -1+√2}

C)
Resolvendo a equação
$(x+2)^{2}=4\\
(x+2)\times (x+2)=4\\
x^{2}+2x+2x+4=4\\
x^{2}+4x+4-4=0\\
x^{2}+4x=0$

A= 1, B= 4, C= 0
$x=\frac{-b\pm \sqrt{b^{2}-4\times a \times c} }{2\times a}\\\\ x=\frac{-4\pm \sqrt{4^{2}-4\times 1 \times 0} }{2\times 1}\\\\ x=\frac{-4\pm \sqrt{16-0} }{2}\\\\ x=\frac{-4\pm \sqrt{16} }{2}\\\\ x=\frac{-4\pm 4 }{2}\\\\ x'= \frac{-4+4}{2}=\frac{0}{2}= \boxed{0} \\\\ x''=\frac{-4-4}{2}= \frac{-8}{2}= \boxed{-4}\\\\ S=\{-4\:,\:0\}$

S= {-4, 0}

$D\'uvidas?\\ Escreva\ \:nos\ \:coment\'arios.$
Espero ter ajudado
Bons estudos

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