Question

Dacă x1,x2 sunt rădăcinile exuației, 2$x^{2}$-8x-10=0, fără a rezolva ecuația, calculați:
a) x1+x2
b) x1 · x2
c) x1² +x2²
d) $\frac{2x1+3}{4x1+5}$ + $\frac{2x2+3}{4x2+5}$
e)$\frac{5x1+2}{3x2+4} + \frac{5x2+2}{3x1+4}$

Answer 1

2x²-8x-10=0
a=2
b=-8
c=-10
x₁+x₂=-b/a=-(-8)/2=4
x₁×x₂=c/a=-10/2=-5
x₁²+x₂²=(x₁+x₂)²-2x₁×x₂=4²-2(-5)=16+10=26

(2x₁+3)/(4x₁+5)+(2x₂+3)/(4x₂+5)=
=(2x₁+3)(4x₂+5)+(2x₂+3)(4x₁+5)/(4x₂+5)/(4x₁+5)=
=(8x₁x₂+12x₂+10x₁+15+8x₁x₂+12x₁+10x₂+15)/(16x₁x₂+20x₁+20x₂+25)=
=(16x₁x₂+22x₂+22x₁+30)/(16x₁x₂+20x₁+20x₂+25)=
=[16x₁x₂+22(x₂+x₁)+30)]/[16x₁x₂+20(x₁+x₂)+25]=
=[16×(-5)+22×4+30]/[16×(-5)+20×4+25]=
=(-80+88+30)/(-80+80+25)=
=38/25


(5x₁+2)/(3x₂+4)+(5x₂+2)/(3x₁+4)=
=(5x₁+2)(3x₁+4)+(5x₂+2)(3x₂+4)/(3x₂+4)/(3x₁+4)=
=(15x₁²+6x₁+20x₁+8+15x₂²+6x₂+20x₂+8)/(9x₁x₂+12x₁+12x₂+16)=
=[15(x₁²+x₂²)+26(x₂+x₁)+16]/[9x₁x₂+12(x₁+x₂)+16]=
=(15×26+26×4+16)/[9×(-5)+12×4+16]=
=(390+104+16)/[(-45)+48+16]=
=510/19