Question

Simplify PLEASE HELP ASAP

Answer 1

$\qquad\qquad\huge\underline{{\sf Answer}}$

Let's evaluate ~

$\qquad \sf \dashrightarrow \: \sqrt{ {x}^{2} - 6x + 9 }$

$\qquad \sf \dashrightarrow \: \sqrt{ {x}^{2} - 3x - 3x+ 9 }$

$\qquad \sf \dashrightarrow \: \sqrt{ {x}^{}(x - 3) - 3(x - 3) }$

$\qquad \sf \dashrightarrow \: \sqrt{ (x - 3)(x - 3) }$

$\qquad \sf \dashrightarrow \: \sqrt{ (x - 3) {}^{2} }$

$\qquad \sf \dashrightarrow \: { (x - 3) {}^{} }$

Now, we have been given that value of x :

$\qquad \sf \dashrightarrow \:x < 3$

So, let's plug the value of x as 3 in the given expression ~

$\qquad \sf \dashrightarrow \:3 - 3$

$\qquad \sf \dashrightarrow \:0$

Therefore, we can conclude that :

$\qquad \sf \dashrightarrow \: \sqrt{ {x}^{2} - 6x + 9 } < 0$

Value of the expression should be less than 0

Answer 2

Answer:

$0 < x < 6$

Step-by-step explanation:

This question asks to solve an inequality which can be written as:

$\sqrt{x^2-6x+9} < 3$

Solve for x:

$\sqrt{x^2-6x+9} < 3\\x^2-6x+9 < 9\\x^2-6x < 0\\x(x-6) < 0$

From here we can deduce that the inequality is $0$ when:

$x = 0$ or $x = 6$

We can write the solution as:

$0 < x < 6$