Question

1) Determine o conjunto solução das equações abaixo: a) x² 8x + 12 = 0 b) x² + 2x8=0 c) x2 - 5x + 8 = 0 d) e) x² - 4x 5=0 f) X² + x + 12 = 0 2x²8x+8 = 0 1 ) Determine o conjunto solução das equações abaixo : a ) x² 8x + 12 = 0 b ) x² + 2x8 = 0 c ) x2 - 5x + 8 = 0 d ) e ) x² - 4x 5 = 0 f ) X² + x + 12 = 0 2x²8x + 8 = 0​

Answer 1

$a)x^2 - 8x + 12 = 0\\\\x = \frac{-b +-\sqrt{b^2 - 4ac }}{2a} \\\\a = 1; b = -8 ; c = 12\\\\x = \frac{-(-8) +- \sqrt{(-8)^2-4 .1.12} }{2.1} = > \\x= \frac{8+-\sqrt{64 -48} }{2} = > \\x = \frac{8+- \sqrt{16} }{2} = > \\x = \frac{8+-\sqrt{4^2} }{2} \\\\x^1 = \frac{8+4}{2} = \frac{12}{2} = > x^1= 6\\\\x^2 = \frac{8-4}{2} =\frac{4}{2}= > x^2 = 2\\ \\s=[2,6]$

$b)x^2+2x-8=0\\\\a=1; b= 2; c=-8\\\\x=\frac{-b+- \sqrt{b^2 - 4ac} }{2a} \\x=\frac{-2+- \sqrt{2^2 - 4.1.-8} }{2} \\\\x=\frac{-2+- \sqrt{4 +32} }{2} \\\\x=\frac{-2+- \sqrt{36} }{2}\\ \\x=\frac{-2+- \sqrt{6^2 } }{2}\\ \\x^1= \frac{-2+6}{2}= > \frac{4}{2} = > x^1 = 2\\\\x^2 = \frac{-2-6}{2}= > \frac{-8}{2} = > x^2 = -4\\ \\s=[-4,2]$

$c) x^2-5x+8 = 0\\\\a =1; b=-5; c = 8\\\\x=\frac{-b+- \sqrt{b^2 - 4ac} }{2a} \\\\x=\frac{-(-5)+- \sqrt{(-5)^2 - 4.1.8} }{2} \\\\x=\frac{5+- \sqrt{25 - 32} }{2} \\\\x=\frac{5+- \sqrt{-7*} }{2}$

* = quando temos raiz quadrada negativa, não temos raízes reais

$d) x^2-4x-5 = 0\\\\a=1; b=-4; c=-5\\\\x=\frac{-b+- \sqrt{b^2 - 4ac} }{2a} \\x=\frac{-(-4)+- \sqrt{(-4)^2 - 4.1.-5} }{2} \\\\x=\frac{4+- \sqrt{16 +20} }{2} \\x=\frac{4+- \sqrt{36} }{2} \\x=\frac{4+- \sqrt{6^2} }{2} \\\\x^1 = \frac{4+6}{2} = > \frac{10}{2}= > x^1 = 5\\\\ x^2 = \frac{4-6}{2} = > \frac{-2}{2}= > -1\\ \\s= [-1,5]$

$e)x^2 +x+12 = 0\\\\a=1; b=1; c=12\\\\x=\frac{-b+- \sqrt{b^2 - 4ac} }{2a} \\x=\frac{-1+- \sqrt{1^2 - 4.1.12} }{2} \\\\x=\frac{-1+- \sqrt{1 - 48} }{2} \\\\x=\frac{-1+- \sqrt{-47*} }{2}$

* = quando temos raiz quadrada negativa, não temos raízes reais

$f) 2x^2 - 8x + 8 = 0\\\\a=2; b= -8; c = 8\\\\x=\frac{-b+- \sqrt{b^2 - 4ac} }{2a}\\x=\frac{-(-8)+- \sqrt{(-8)^2 - 4.2.8} }{2.2}\\\\x=\frac{8+- \sqrt{64 -64} }{4}\\\\x=\frac{8+- \sqrt{0*} }{4}\\\\\sqrt{0} = > x^1 = x^2 = > \\ \\x=\frac{8 }{4} = > x = 2\\\\s=[2]$

* = quando temos raiz quadrada igual a zero, o x1 e x2 serão iguais, cortando a raiz na equação.