1. anonymous

why wouldn't it be x = 5, not extraneous ?

2. Nnesha

3. anonymous

i know, I should've subtracted five, and then square booth sides

4. Nnesha

wrong.

5. anonymous

I had just gone straight to squaring the whole equation?

6. Nnesha

5 is under the square root so you can't move it to the right without removing square root

7. anonymous

then subtract?

8. anonymous

lemme resolve

9. anonymous

do you still want a picture of my work?

10. Nnesha

yep

11. anonymous

this is how i got the x = 5, not extraneous

12. anonymous

re solving atm

13. anonymous

but i did square first there ^

14. Nnesha

well well well. first step is to divide both side by 2

15. Nnesha

then take square root both sides

16. anonymous

cause of the two infront of the radical?

17. Nnesha

yep right you need to transfer all numbers that is outside the radical sign to the right side that's always going to be First step

18. Nnesha

to take square root you need just radical sign at left side

19. anonymous

|dw:1438802258254:dw|

20. Nnesha

2/2 = ?

21. anonymous

|dw:1438802316942:dw|

22. Nnesha

it's not 0

23. anonymous

one,

24. Nnesha

yes right $\huge\rm \sqrt{x-5}=1$ now solve for x

25. anonymous

ohhhhh!

26. anonymous

x - 5 = 1 x = 6 not extraneous

27. Nnesha

true

28. anonymous

yay! thank you !

29. Nnesha

my pleasure :=)