Question

The number of intergral values $x$ which satisfies $x-3<\sqrt{x+27}$ is

Answer 1

Given $x-3<\sqrt{x+27}$ to be defined
$x+27\ge 0\Rightarrow x\ge -27\cdots \left(1\right)$

$\sqrt{x+27}>x-3$
Now $\sqrt{x+27}\ge 0$

$\therefore \text{If}x-3<0\Rightarrow x<3$
$\Rightarrow x\in [-27,3)\cdots \left(2\right)$

$\text{If}x-3\ge 0\Rightarrow x\ge 3$
$\sqrt{x+27}>x-3$
squaring both the sides
$\Rightarrow x+27>(x-3{)}^2\Rightarrow x^2-7x-18<0\Rightarrow (x+2\left)\right(x-9)<0\Rightarrow x\in (-2,9)$
but $x\ge 3$
$\therefore x\in [3,9)\cdots \left(3\right)$

From $\left(1\right)$ and $\left(2\right)$
$x\in [-27,9)$
Hence there are total $36$ integral values