• Matéria: Matemática
  • Autor: Paamorim
  • Perguntado 9 anos atrás

Sendo x¹ e x² raízes da equação 2x²-7x + 4 = 0 , determine o valor de x1² + x2²

Respostas

respondido por: ptpp95
8

Fórmula de Báskaras:

 

Delta= b²-4.a.c

 

D= 49-4.(2).(4)

 

D=49-32

 

D=<var>\sqrt{17}</var>

 

-(b)+-<var>\sqrt{17}</var>/2.(a)

 

7+-<var>\sqrt{17}</var>/4

 

x'= 7+17/4=12

 

x"=7-17/4=-10/4

respondido por: justinianolima
2

Δ=b²-4ac

Δ=(-7)²-4*2*4

Δ=49-32

Δ=17

<var>x^1=\frac{-x+\sqrt{17}}{2*2}=\frac{-x+\sqrt{17}}{4} \\x^2=\frac{-x-\sqrt{17}}{2*2}=\frac{-x-\sqrt{17}}{4}</var>

Achada as raizes temos:

<var>(\frac{-(-7)+\sqrt{17}}{4})^2+(\frac{-(-7)-\sqrt{17}}{4})^2= (\frac{7+\sqrt{17}}{4})^2+(\frac{7-\sqrt{17}}{4})^2=\\ \frac{(-7)^2+2(-7)(\sqrt{17})+\sqrt{17}^2}{4^2}+\frac{(-7)^2-2(-7)(\sqrt{17})+\sqrt{17}^2}{4^2}=\\ \frac{2*49+2*17}{16}=\frac{98+34}{16}=\frac{132}{16}=\frac{33}{4}=8,25</var>

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