Question

Let $y=\sqrt{\frac{(x+1)(x-3)}{x-2}}.$
Find all the real values of x for which y takes real values.

Answer 1

$y=\sqrt{\frac{(x+1)(x-3)}{x-2}}.$

For $y$ to be real, the term under the square root should be greater or equal to 0

$\frac{(x+1)(x-3)}{x-2}\ge 0$

Case I: For this, the numerator should be greater than equal to 0 and denominator should be greater than 0

i.e $(x+1)(x-3)\ge 0$ and $x-2>0$

$\Rightarrow x\in (-\infty ,-1)\cup [3,\infty ]$ and $x>2$

So, the solution set is $[3,\infty )$

Case II: For this, the numerator should be less than equal to 0 and denominator should be less than 0

i.e $(x+1)(x-3)\le 0$ and $x-2<0$

$\Rightarrow x\in [-1,3]$ and $x<2$

So, the solution set is $[-1,2)$

Hence, the domain is $[-1,2)\cup [3,\infty )$