Question

Fie ecuația ( m+3 ) x² -2(m-1)x+m-2=0

b) Sa se determine m aparține Nr. reale pentru care soluțiile reale ale ecuatiei verifică relația... ( imaginea de mai sus )

Urgent !​

Answer 1

Răspuns:

$(m+3)x^2-2(m-1)x+m-2=0; a=m+3;b=-2(m-1);c=m-2$

$S=x1+x2=\frac{-b}{a}=\frac{2m-2}{m+3}$

$P=x1x2=\frac{c}{a}=\frac{m-2}{m+3}$

$2(x1+x2)-5x1x2=\frac{-2m+5}{2}$

$2(\frac{2m-2}{m+3} )-5(\frac{m-2}{m+3})=\frac{-2m+5}{2}$

$\frac{4m-4-5m+10}{m+3} =\frac{-2m+5}{2}$

$\frac{-m+6}{m+3}=\frac{-2m+5}{2}$

$2(-m+6)=(m+3)(-2m+5)$

$-2m+12=-2m^2+5m-6m+15$

$-2m^2+m+3=0$

$-2m^2-2m+3m+3=0$

$-2m(m+1)+3(m+1)=0$

$(-2m+3)(m+1)=0$

$-2m+3=0=>m=\frac{3}{2}$

$m+1=0=>m=-1$