Question

Укажите решение квадратного неравенства:
2x^2-3x+1>=0

Answer 1

$2x^2-3x+1 \geq 0\\\\ D=(-3)^2-4*2*1=9-8=1\\\\ x_{1,2}=\frac{-(-3)\pm\sqrt{1}}{2*2}=\frac{3\pm1}{4}\\\\ x_1=1\ \ \ \ x_2=\frac{1}{2}\\\\ 2*(x-1)*(x-\frac{1}{2}) \geq 0\\\\ (x-1)*(x-\frac{1}{2}) \geq 0\\\\ +++++++[\frac{1}{2}]---------[1]++++++++\ \textgreater \ x\\\\ x\in(-\infty;\ \frac{1}{2}]\cup[1;\ +\infty)$