Question

x²+2(x+1)=5

alguém me ajuda, não consigo entender.

Answer 1

Resposta:

Explicação passo-a-passo:

x²+2(x+1)=5

x²+2x+2=5

x²+2x-3=0

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+2x-3=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=2~e~c=-3\\\\\Delta=(b)^{2}-4(a)(c)=(2)^{2}-4(1)(-3)=4-(-12)=16\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(2)-\sqrt{16}}{2(1)}=\frac{-2-4}{2}=\frac{-6}{2}=-3\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(2)+\sqrt{16}}{2(1)}=\frac{-2+4}{2}=\frac{2}{2}=1\\\\S=\{-3,~1\}$

Answer 2

Resposta:

$primeiro~aplique~a~propriedade~distributiva~em~2(x+1)\\ \\x^{2}+2x+2=5\\ \\ x^{2}+2x+2-5=0\\ \\x^{2}+2x-3=0\\ \\equacao~do~segundo~grau\\ \\formula~de~Bhaskara\\ \\ a=1\\b=2\\c=-3\\ \\ delta=b^{2}-4.a.c\\ \\delta=2^{2} -4.1.(-3)\\ \\delta=4-(-12)=4+12\\ \\ delta=16$$\\ \\x=\frac{-b+-\sqrt{delta}}{2.a}\\\\x=\frac{-2+-\sqrt{16}}{2}\\\\x=\frac{-2+-4}{2}\\ \\x'=\frac{-2-4}{2}=\frac{-6}{2}=-3\\ \\x''=\frac{-2+4}{2}=\frac{2}{2}=1\\ \\ S~~(-3;~~1)$