Question

2,3 (найти область определения)

Answer 1

$y(x)=\sqrt{2x^2-5x-3}$

$2x^2-5x-3 \geq 0\\\\ D=(-5)^2-4*2*(-3)=25+24=49=7^2\\\\ x_{1,2}=\frac{-(-5)\pm\sqrt{7^2}}{2*2}=\frac{5\pm7}{4}\\\\ x_1=3\ \ \ x_2=-\frac{1}{2}\\\\ a(x-x_1)(x-x_2)=0\\\\ 2*(x-3)*(x-(-\frac{1}{2})) \geq 0\\\\ (x-3)*(x-(-0.5)) \geq 0\\\\ +++++++[-0.5]------[3]+++++++++\ \textgreater \ x\\\\ x\in(-\infty;\ -0.5)\cup(3;\ +\infty)$