Question

obtenha quando existir, os zeros das funções dadas por

a) g(x) = x^2 + 3 + 2
b) g(x) = 2x^2 + x + 1
c) 9x^2 + 6x + 1

Answer 1

Resposta:

Os zeros da função são obtidos igualando as funções a zero.

a)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+3x+2=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=3~e~c=2\\\\\Delta=(b)^{2}-4(a)(c)=(3)^{2}-4(1)(2)=9-(8)=1\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(3)-\sqrt{1}}{2(1)}=\frac{-3-1}{2}=\frac{-4}{2}=-2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(3)+\sqrt{1}}{2(1)}=\frac{-3+1}{2}=\frac{-2}{2}=-1\\\\S=\{-2,~-1\}$

b)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~2x^{2}+1x+1=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=2{;}~b=1~e~c=1\\\\\Delta=(b)^{2}-4(a)(c)=(1)^{2}-4(2)(1)=1-(8)=-7\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(1)-\sqrt{-7}}{2(2)}=\frac{-1-i\sqrt{7}}{4}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(1)+\sqrt{-7}}{2(2)}=\frac{-1+i\sqrt{7}}{4}\\\\\\S=\{\frac{-1-i\sqrt{7}}{4},\frac{-1+i\sqrt{7}}{4}\}$

c)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~9x^{2}+6x+1=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=9{;}~b=6~e~c=1\\\\\Delta=(b)^{2}-4(a)(c)=(6)^{2}-4(9)(1)=36-(36)=0\\\\x^{'}=x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(6)+\sqrt{0}}{2(9)}=\frac{-6+0}{18}=\frac{-6}{18}=-\frac{1}{3} \\\\S=\{-\frac{1}{3} \} ~ra\'{i}z~dupla$