Question

f:R->R
f(x)=x²+6x-5
---------------------
x²1+x²2 = ???
x³1+x³2=????

REPEDE VA ROG !!!!

Answer 1

Explicație pas cu pas:

Folosim relatiile lui Viete

pentru ecuatia:

f(x)=x²+6x-5

x₁+x₂=-6

x₁*x₂=-5

x²₁+x²₂=(x₁+x₂)²-2*x₁*x₂=(-6)²+2*5=36+10=46

x₁³+x₂³=

(x₁+x₂)(x₁²-x₁*x₂+x₂²)=-6((x₁+x₂)²-3*x₁*x₂)=-6((-6)²+15)=-6(36+15)=

=-6*51=-306

Bafta!

Answer 2

$\mathrm{f(x) = x^2+6x-5} \\ \\ \mathrm{x_1^2+x_2^2 = (x_1+x_2)^2-2x_1x_2 = (-6)^2-2\cdot (-5) = 46} \\ \\ \left\{\begin{array}{II} \mathrm{x_1^2+6x_1-5 = 0\Big|\cdot x_1} \\ \mathrm{x_2^2+6x_2-5 = 0\Big|\cdot x_2} \end{array} \Rightarrow\left\{\begin{array}{II} \mathrm{x_1^3+6x_1^2-5x_1 = 0} \\ \mathrm{x_2^3+6x_2^2-5x_2 = 0} \end{array} \Rightarrow$

$\Rightarrow \mathrm{x_1^3+x_2^3 +6(x_1^2+x_2^2)-5(x_1+x_2) = 0} \Rightarrow \\ \\ \Rightarrow \mathrm{x_1^3+x_2^3 = 5(x_1+x_2) - 6(x_1^2+x_2^2) = -30 -6\cdot 46 = -306 }$