Question

Determine a soma dos quadrados dos inversos das raizes de :

Answer 1

$( \frac{x}{4} + \frac{1}{3} )( \frac{x}{6} - \frac{1}{6} ) = 0 \\ \frac{ {x}^{2} }{24} - \frac{x}{24} + \frac{x}{18} - \frac{1}{18} = 0$

$\frac{3 {x}^{2} - 3x + 4x - 4 }{72} = 0 \\ 3 {x}^{2} + x - 4 = 0$

∆=b²-4.a.c

∆=1²-4.3.(-4)

∆=1+48=49

x= (-b±√∆)/2a

x=(-1±√49)/2.3

x=(-1±7)/6

x₁=(-1+7)/6=6/6=1

x₂=(-1-7)/6=-8/6=-4/3

1/x₁² +1/x₂²

$\frac{1}{ {1}^{2} } + \frac{1}{ {( - \frac{4}{3} )}^{2} } = 1 + \frac{1}{ \frac{16}{9} } \\ = 1 + \frac{9}{16} = \frac{16 + 9}{16} = \frac{25}{16}$