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 --> \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\).
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