Question

sq root x - 2 + 5 = 6

Answer 1

\[\Large \sqrt{x-2}+5 = 6\]
this?

Answer 2

yes

Answer 3

I want to learn the steps

Answer 4

Yup :)
First, put everything OUTSIDE THE SQUARE ROOT SIGN to the right-side of the equation.... For instance, this +5 here...
\[\Large \sqrt{x-1}\color{red}{+5}= 6\]
How to do that?

Answer 5

Crud, sorry, typo....
\[\Large \sqrt{x-2}\color{red}{+5}= 6\]

Answer 6

So you can do that by subtracting 5 from both sides...
\[\Large \sqrt{x-2} \ {+ \ 5}\color{red}{-5}= 6\color{red}{-5}\]

Answer 7

Leaving you with...
\[\Large \sqrt{x-2}=\color{red}1\]
can you do it from here?

Answer 8

No

Answer 9

Well, the most natural thing to do would be to square both sides, as squaring cancels out the radical... let me explain...
\[\Large (\sqrt{a})^2 = a\]

Catch me so far?

Answer 10

Yes

Answer 11

Actually, no, that makes no sense

Answer 12

If you square a radical, you just remove the radical.

Answer 13

Or, if you prefer it worded this way, if you raise a radical to an exponent 2, the exponent and the radical would "disappear" or cancel each other out.

Answer 14

okay:) ready for the next step please

Answer 15

Okay, so squaring both sides....
\[\Large (\sqrt{x-2})^2 = 1^2\]
Now, \(\Large 1^2 = 1\)
so it simplifies into...
\[\Large (\sqrt{x-2})^2 = 1\]
What does the left side simplify into?

Answer 16

x^2?

Answer 17

No... I just told you, squaring would just remove the radical \(\Large \sqrt{\color{white}{skjassj}}\) sign

Answer 18

So nothing?

Answer 19

Okay... back to square one...
If we square a radical, both the square (the 2 exponent) and the radical "disappear"

Case in point
\[\Large (\sqrt3)^2 = 3\]\[\Large (\sqrt{a+b+c})^2 = a+b+c\]

and so on.

So, to what does the left side of this equation simplify into...
\[\Large (\sqrt{x-2})^2 = 1\] ?

Answer 20

2?

Answer 21

this makes no sense okay? it really doesn't

Answer 22

im sorry

Answer 23

Perhaps we need a more thorough explanation? :)

Answer 24

okay, we have a radical expression... with something, anything, inside the radical...
\[\Large \left.\sqrt{<something>}\right.\]

If we square it... or raise it to an exponent 2,
\[\Large \left(\sqrt{<something>}\right)^2\]

Then the \(\large 2\) and the radical sign \(\Large \sqrt{\color{white}{text}}\)

would cancel each other out, leaving you with whatever was inside the radical...

\[\Large \left(\sqrt{<something>}\right)^2=<something>\]