Respuesta :

nabeelahmed1721

Answer:

[tex]0 < x < 6[/tex]

Step-by-step explanation:

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

[tex]\sqrt{x^2-6x+9} < 3[/tex]

Solve for x:

[tex]\sqrt{x^2-6x+9} < 3\\x^2-6x+9 < 9\\x^2-6x < 0\\x(x-6) < 0[/tex]

From here we can deduce that the inequality is [tex]0[/tex] when:

[tex]x = 0[/tex] or [tex]x = 6[/tex]

We can write the solution as:

[tex]0 < x < 6[/tex]

DᴀʀᴋPᴀʀᴀᴅᴏx

[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]

Let's evaluate ~

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 6x + 9 } [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 3x - 3x+ 9 } [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{}(x - 3) - 3(x - 3) } [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ (x - 3)(x - 3) } [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ (x - 3) {}^{2} } [/tex]

[tex]\qquad \sf  \dashrightarrow \: { (x - 3) {}^{} } [/tex]

Now, we have been given that value of x :

[tex]\qquad \sf  \dashrightarrow \:x < 3[/tex]

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

[tex]\qquad \sf  \dashrightarrow \:3 - 3[/tex]

[tex]\qquad \sf  \dashrightarrow \:0[/tex]

Therefore, we can conclude that :

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 6x + 9 } < 0[/tex]

Value of the expression should be less than 0

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