Question

URGENT ,VA ROG MULT
Sa se determine solutiile reale ale ecuatiei f la puterea a doua(x) +2f (x) -3=0 unde f:R-> R ,f (x)=2x-1

Answer 1

$f(x) = 2x-1 \\ \\ f^2(x) + 2f(x)-3 = 0 \Rightarrow \Big[f(x)\Big]^2+2f(x) - 3 = 0 \\ \Delta = 2^2-4\cdot 1\cdot (-3) \Rightarrow \Delta = 4 + 12 \Rightarrow \Delta = 16 \\ \\ \Rightarrow f(x)_{1,2} = \dfrac{-2\pm\sqrt{16} }2\cdot 1} \Rightarrow f(x)_{1,2} = \dfrac{-2\pm4}{2} \\ \\ f(x)_1 = 1,\quad f(x)_2 = -3 \\ \\ \boxed{1}\quad f(x)_1 = 1 \Rightarrow 2x-1 = 1 \Rightarrow 2x = 2 \Rightarrow x_1 = 1 \\ \\ \boxed{2} \quad f(x)_2 = -3 \Rightarrow 2x-1 = -3 \Rightarrow 2x = -2 \Rightarrow x_2 = -1$

$\Rightarrow S = \Big\{-1,1\Big\}$