Question

Ecuatia x^2 - x + m = 0 are solutiile x1 si x2 .Sa se determine nr. real m pentru care 1/(x1+1)+1/(x2+1)=-3/4

Answer 1

Prelucram mai intai ecuatia ta
$\frac{1}{x1+1}+\frac{1}{x2+1}=\frac{x2+1+x1+1}{(x1+1)(x2+1)=\frac{x1+x2+2}{x1x2+x1+x2+1}=\frac{S+2}{S+P+1}$ Unde am notat S=x1+x2, si P=x1x2

Tu stii ca solutiile x1 si x2 au urmatoarele formule
$x1=\frac{-b+\sqrt{\Delta}}{2a}$
$x2=\frac{-b-\sqrt{\Delta}}{2a}$
unde
$\Delta=b^(2)-4ac$
Atunci hai sa vedem cat este suma si produsul lor
$S=x1+x2=\frac{-b+\sqrt{\Delta}}{2a}+\frac{-b-\sqrt{\Delta}}{2a}=\frac{-2b}{2a}=-\frac{b}{a}$
In cazul nostru concret: a=1, b=-1 si c=m
Deci avem
$S=-\frac{1}{-1}=1$
Produsul este
$P=x1x2=\frac{-b+\sqrt{\Delta}}{2a}*\frac{-b-\sqrt{\Delta}}{2a}=\frac{1}{4a^{2}}(-b+\sqrt{Delta})(b-\sqrt{Delta})=\frac{1}{4a^{2}}(b^{2}-\Delta)=\frac{1}{4a^{2}}(b^{2}-b^{2}+4ac)=\frac{4ac}{a^{2}}=\frac{c}{a}$
In cazul nostru concret
$P=\frac{m}{1}=m$
Relatiile pentru S si P sunt universale, pentru orice ecuatie de gradul 2
Inlocuim in ecuatia de mai sus
$\frac{S+2}{S+P+1}=\frac{1+2}{1+m+1}=\frac{3}{m+2}=\frac{-3}{4}\Rightarrow m+2=-4\Rightarrow m=-6$

Answer 2

x² - x + m = 0      x1 = (1 +√Δ)/2    (Δ= 1- 4m)      x1 + 1 = (3+√Δ)/2  
1/(3+√Δ)/2 = 2( 3-√Δ )/(9 - Δ)
x2 = (1 - √Δ)/2    x2 + 1 = (3 - √Δ )/2    1/ (3-√Δ)2 = 2(3+√Δ) /(9 - Δ)
(3- √Δ+ 3+√Δ) / (9 -Δ) = -3/4
6 / (9-Δ) = -3/4      9 - Δ  = -8   Δ = 17     1 - 4m = 17  4m = - 16    m = - 4