Question

Fie x1 si x2 radacinile reale ale ecuatiei 2x^2+5x+1=0. Atunci x1^2(x2+1)+x2^2(x1+1) are solutia?

Answer 1

Vom scrie a doua relatie in functie de suma si produsul radacinilor primei ecuatii:
$x_{1} ^{2}( x_{2}+1 ) +x_{2} ^{2}( x_{1}+1 )= x_{1} ^{2} x_{2}+x_{1} ^{2}+x_{2} ^{2} x_{1}+x_{2} ^{2}= \\ x_{1} x_{2}( x_{1}+ x_{2})+x_{1} ^{2}+ x_{2} ^{2} =x_{1} x_{2}( x_{1}+ x_{2})+ (x_{1}+ x_{2})^{2}-2x_{1} x_{2}$

Stim ca [tex] x_{1}+ x_{2}= -b/a=-5/2 \\ x_{1} x_{2}=c/a=1/2 [/tex]

Inlocuim in relatia obtinuta mai sus:
$\frac{1}{2}( \frac{-5}{2} )+( \frac{-5}{2})^{2} -2( \frac{1}{2} )=$-5/4+25/4-1=20/4-1=4

Answer 2

$2x^2+5x+1=0\\ \\ 2x_{1}^2+5x_1+1 = 0 \Rightarrow 2x_1^2 = -5x_1-1 \Rightarrow x_1^2 = -\dfrac{5x_1+1}{2} \\ \\ 2x_{2}^2+5x_2+1 = 0 \Rightarrow 2x_2^2 = -5x_2-1 \Rightarrow x_2^2 = -\dfrac{5x_2+1}{2}\\ \\ \\ \\x_1^2(x_2+1)+x_2^2(x_1+1) = \\ \\ = -\dfrac{5x_1+1}{2} \cdot(x_2+1)+\Big(-\dfrac{5x_2+1}{2}\Big)(x_1+1) \\ \\ =- \dfrac{(5x_1+1)(x_2+1)}{2} - \dfrac{(5x_2+1)(x_1+1)}{2} =$

$= -\dfrac{(5x_1+1)(x_2+1)+(5x_2+1)(x_1+1)}{2} = \\ \\ = -\dfrac{5x_1x_2+5x_1+x_2+1+5x_2x_1+5x_2+x_1+1}{2} = \\ \\ = -\dfrac{5x_1x_2+5x_2x_1+5(x_1+x_2)+x_1+x_2+2}{2} = \\ \\ = -\dfrac{5x_1x_2+5x_1x_2+5(x_1+x_2)+(x_1+x_2)+2}{2} = \\ \\ = - \dfrac{10x_1x_2 +6(x_1+x_2)+2}{2} = \\ \\ = -\dfrac{10\cdot\dfrac{1}{2}+6\cdot \dfrac{-5}{2}+2}{2} = \\ \\ = -\dfrac{5-15+2}{2} = \\ \\ = -\dfrac{-8}{2} = \\ \\ = -(-4) = \\ \\ = \boxed{4}$