Question

Se consideră funcţiile f , g :R→R, f (x) = x2 − 3x + 2, g(x) = 2x −1. Să se rezolve ecuaţia
( f compus g)(x) = 0 .

Answer 1

   
[tex]\displaystyle \\ f(x)=x^2-3x+2 \\ g(x)=2x-1 \\ f \circ g = 0 \\ (2x-1)^2-3(2x-1)+2=0 \\ 4x^2 -4x + 1 - 6x + 3 + 2 = 0 \\ 4x^2 -10x + 6 = 0 ~~~~~|:2\\ 2x^2 - 5x +3 =0 \\ \\ x_{12}= \frac{-b\pm \sqrt{b^2-4ac} }{2a} =\frac{5\pm \sqrt{25-24} }{4}= \frac{5 \pm 1}{4} \\ \\ x_1 = \frac{5 +1}{4} =\frac{6}{4} =\boxed{\frac{3}{2} } \\ \\ x_2= \frac{5 -1}{4} =\frac{4}{4} =\boxed{1 } [/tex]