Answer:- x-9=±9
Explanation:-
Given:-
[tex](x-9)^2=81\\\Rightarrow(x-9)^2-81=0\\\Rightarrow(x-9)^2-9^2=0\\\Rightarrow[(x-9)+9][(x-9)-9]=0.............by\ identity\ a^2-b^2=(a+b)(a-b)[/tex]
[tex]\Rightarrow[(x-9)+9]=0\ or\ [(x-9)-9]=0\\\Rightarrow(x-9)=-9\ or\ (x-9)=9[/tex]
⇒x-9=±9
Thus on taking square root ob both sides of [tex](x-9)^2=81[/tex], we get
[tex]\sqrt{(x-9)^2}= \sqrt{81}[/tex]
[tex]\\\Rightarrow(x-9)=\sqrt{9^2}[/tex]
⇒x-9=±9
The equation results from taking the square root of both sides of (x – 9)2 = 81 are x – 9 = ±9.
We have to determine
Which equation results from taking the square root of both sides of (x – 9)2 = 81?
The square root of a number is a value, which on multiplication by itself gives the original number.
Equation; [tex]\rm (x - 9)^2 = 81[/tex]
Taking square root on both sides
[tex]\rm \rm (x - 9)^2 = 81\\\\ \rm (x - 9)^2 = 9^2\\\\\\sqrt{(x - 9)^2 }= \sqrt{9^2}\\\\(x-9)=\pm 9[/tex]
Hence, the equation results from taking the square root of both sides of (x – 9)2 = 81 are x – 9 = ±9.
To know more Square root property click the link given below,
https://brainly.com/question/1675383
