Question

Could somebody help me solve this please ?

√13-N = √N-7

Answer 1

@sourwing could you please help me ?

Answer 2

what are the roots and what are not the roots?

Answer 3

is 13-N and N-7 all included inside the the square roots ?

Answer 4

yes

Answer 5

\(\Large\color{blue}{ \sf \sqrt{13-N}= \sqrt{N-7} }\) square both sides


\(\Large\color{blue}{ \sf 13-N= N-7 }\) add N to both sides
\(\Large\color{blue}{ \sf ~~~+N~~~~~+N }\)


\(\Large\color{blue}{ \sf 13= 2N-7 }\) add 7 to both sides
\(\Large\color{blue}{ \sf +7~~~+7 }\)


\(\Large\color{blue}{ \sf 20= 2N }\) divide both sides by 2, ...
\(\Large\color{blue}{ \sf 10= N }\)

Answer 6

you're the bomb.com can i ask you another question? @SolomonZelman

Answer 7

Thx for calling me that, although I am by far not the best mathematician in OS, closer to the worse one....

Answer 8

Yeah, sure, go ahead and ask.

Answer 9

√2x-18 = √x-5

Answer 10

This time, I am going to tell you what to do, and you will do the steps I tell you for me. OK ?

Answer 11

okay

Answer 12

\(\Large\color{blue}{ \sf \sqrt{2x-18} =\sqrt{x-5} }\)

1) Square both sides
2) subtract x from both sides
3) add 18 to both sides

tell me the answer (I would prefer you to show your steps in here) .

Answer 13

this might sound dumb but how do you square the sides ? is that when you multiply them by two ?

Answer 14

I am going to show you ....

once you know that \(\Large\color{blue}{ \bf \sqrt{a} =\sqrt{a} }\) you also know that

\(\Large\color{blue}{ \bf (\sqrt{a})^{~2} =(\sqrt{a})^{~2} }\) and therefore you know that \(\Large\color{blue}{ \bf a =a }\)

when you square both sides, you are simply raising each side to the second power, just like i did in the first problem.

Answer 15

√2x-18^2=√x-5^2 like this??

Answer 16

when you raise a square root of a number to a second power, the root cancels right?