Question

1)Resolva as equações. 20 ponto pra que responde
a)(x+5) ² =16
b)(x+5) ² =9
c)(x+5) ²=4
d)(x+5) ²=1
e)(x+5) ²=0

2)Explique por que a equação abaixo não tem raízes real .
(3x-1)²=-4

Answer 1

a)
$(x+5).(x+5) = 16$
Vamos fazer a distributiva, multiplique x por 5 e x, e 5 por x e 5 irá ficar assim:
$x^{2} + 5x + 5x + 25 = 16\\ x^{2} + 10x + 25 - 16 = 0\\ x^{2} + 10x + 9 = 0\\\\ Delta = b^{2} -4.a.c \\Delta = 10^{2} - 4.1.9 \\Delta = 100 - (4.9)\\ Delta = 100 - (+36) \\Delta = 100 -36 \\Delta = 64\\\\ x =\frac{-b +- \sqrt{Delta} }{2.a}\\ x =\frac{-(+10) +- \sqrt{64} }{2.1}\\\\ x =\frac{-10 +- 8}{2}\\\\ x_{1}= \frac{-10+(-8)}{2} = \frac{-10-8}{2} = \frac{-18}{2} = -9\\\\ x_{2}= \frac{-10-(-8)}{2} = \frac{-10+8}{2} = \frac{-2}{2} = -1$
Resposta: -9 e -1

b)
$(x+5).(x+5)=9$
$x^{2} + 5x + 5x + 25 = 9\\ x^{2} + 10x + 25 - 9 = 0\\ x^{2} + 10x +16 = 0\\\\ Delta = b^{2} -4.a.c \\Delta = 10^{2} - 4.1.16 \\Delta = 100 - (4.16)\\ Delta = 100 - (+64) \\Delta = 100 -64 \\Delta = 36\\\\ x =\frac{-b +- \sqrt{Delta} }{2.a}\\ x =\frac{-(+10) +- \sqrt{36} }{2.1}\\\\ x =\frac{-10 +- 6}{2}\\\\ x_{1}= \frac{-10+6}{2} = \frac{-10+6}{2} = \frac{-4}{2} = -2\\\\ x_{2}= \frac{-10-6}{2} = \frac{-10-6}{2} = \frac{-16}{2} = -8$
Resposta: -2 e -8

c)
$(x+5).(x+5)=4\\\\ x^{2} + 5x + 5x + 25 = 4\\ x^{2} + 10x + 25 - 4 = 0\\ x^{2} + 10x +21 = 0\\\\ Delta = b^{2} -4.a.c \\Delta = 10^{2} - 4.1.21 \\Delta = 100 - (4.21)\\ Delta = 100 - (+84) \\Delta = 100 -84 \\Delta = 16\\\\ x =\frac{-b +- \sqrt{Delta} }{2.a}\\ x =\frac{-(+10) +- \sqrt{16} }{2.1}\\\\ x =\frac{-10 +- 4}{2}\\\\ x_{1}= \frac{-10+4}{2} = \frac{-10+4}{2} = \frac{-6}{2} = -3\\\\ x_{2}= \frac{-10-4}{2} = \frac{-10-4}{2} = \frac{-14}{2} = -7$
Resposta: -3 e -7

d)
$(x+5).(x+5)=1\\\\ x^{2} + 5x + 5x + 25 = 1\\ x^{2} + 10x + 25 - 1 = 0\\ x^{2} + 10x +24= 0\\\\ Delta = b^{2} -4.a.c \\Delta = 10^{2} - 4.1.24 \\Delta = 100 - (4.24)\\ Delta = 100 - (+96) \\Delta = 100 -96 \\Delta = 4\\\\ x =\frac{-b +- \sqrt{Delta} }{2.a}\\ x =\frac{-(+10) +- \sqrt{4} }{2.1}\\\\ x =\frac{-10 +- 2}{2}\\\\ x_{1}= \frac{-10+2}{2} = \frac{-10+2}{2} = \frac{-8}{2} = -4\\\\ x_{2}= \frac{-10-2}{2} = \frac{-10-2}{2} = \frac{-12}{2} = -6$
Resposta: -4 e -6

e)
$(x+5).(x+5)=0\\\\ x^{2} + 5x + 5x + 25 = 0\\ x^{2} + 10x + 25 - 0 = 0\\ x^{2} + 10x +25= 0\\\\ Delta = b^{2} -4.a.c \\Delta = 10^{2} - 4.1.25 \\Delta = 100 - (4.25)\\ Delta = 100 - (+100) \\Delta = 100 -100 \\Delta = 0\\\\ x =\frac{-b +- \sqrt{Delta} }{2.a}\\ x =\frac{-(+10) +- \sqrt{0} }{2.1}\\\\ x =\frac{-10 +- 0}{2}\\\\ x_{1}= \frac{-10+0}{2} = \frac{-10+0}{2} = \frac{-10}{2} = -5\\\\ x_{2}= \frac{-10-0}{2} = \frac{-10-0}{2} = \frac{-10}{2} = -5$
Resposta: -5

2)
Não há raizes pois o delta é negativo