Question

Problemas com equações do 2ª grau

Answer 1

Resposta:

Explicação passo-a-passo:

a)

2x²=3x-1

2x²-3x+1=0

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~2x^{2}-3x+1=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=2{;}~b=-3~e~c=1\\\\\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(2)(1)=9-(8)=1\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-\sqrt{1}}{2(2)}=\frac{3-1}{4}=\frac{2}{4}=0,5\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+\sqrt{1}}{2(2)}=\frac{3+1}{4}=\frac{4}{4}=1\\\\S=\{0,5;~1\}$

b)

1-4/x²=3/x => multiplique tudo por x²

x²-4=3x

x²-3x-4=0

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-3x-4=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-3~e~c=-4\\\\\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(1)(-4)=9-(-16)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-\sqrt{25}}{2(1)}=\frac{3-5}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+\sqrt{25}}{2(1)}=\frac{3+5}{2}=\frac{8}{2}=4\\\\S=\{-1,~4\}$