PrayasT wrote:

Hey Bunnel!

I'm a bit confused over here. I do understand that \(\sqrt{x^2} = |x|\). However, we are given that each expression under the square root is greater than or equal to 0. Hence \(\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|\).

So the entity, \(|x-3| ≥ 0\) Hence, isn't the range of x becomes \(x≥3\) from this term?

Let me know where am I going wrong.

Thanks a lot

PrayasT

Absolute value is always non-negative so \(|x-3| ≥ 0\) is true for any value of x. Or in other way, since the square of a number is non-negative then \(\sqrt{(x-3)^2}\) is defined for any value of x.

We get the range of x from \(\sqrt{2 - x}\). The square root to be defined the expression under the square root must be greater than or equal to 0 --> \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\).