Question

Find the domain of definition of the following function.
$y,=\sqrt{\frac{1}{2x^2-5x-3}}.$

Answer 1

$y=\sqrt{\frac{1}{2x^2-5x-3}}$

$(2x^2-5x-3)>0$

$(2x^2-6x+x-3)>0$

$2x(x-3)+(x-3)>0$

$(2x+1)(x-3)>0$

$x>3$

$(2x+1)>0$

$2x>-1$

$x>\frac{-1}{2}$

$x>3$

Also $(2x+1)(x-3)>0$ holds when

$(2x+1)<0$ and $(x-3)<0$

or, $x<-\frac{1}{2}$ and $x<3$

The Domain of $y=(-\infty ,-\frac{1}{2})\cup (3,\infty )$