## anonymous one year ago PLEASE HELP, IM LOST ON THIS QUESTION Given the equation 2Square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution.

1. anonymous

2. anonymous

i did already but i got 2x-5=4

3. anonymous

or would it be 4x-5=4?

4. imqwerty

is this the question - $\sqrt[2]{x-5}=2$??

5. anonymous

yes

6. imqwerty

ok so when u square both the sides $(\sqrt[2]{x+5})^2 =2^2$ what will be the RHS after squaring? :)

7. anonymous

sorry , whats the RHS again?

8. imqwerty

RHS=right hand side :)

9. anonymous

ok thanks it would be

10. imqwerty

right hand side of the equation

11. anonymous

4

12. imqwerty

yes and what wuld be the Left hand side?

13. anonymous

x+5

14. imqwerty

so we get x+5 =4

15. anonymous

yes, now we solve for the variable and get -1 right?

16. anonymous

@imqwerty

17. anonymous

18. imqwerty

yes u get x=-1 :)

19. anonymous

20. imqwerty

ok

21. imqwerty

u need help in??

22. anonymous

well my answer choices are these : x= 6, solution is not extraneous x = 6, solution is extraneous x = 11, solution is not extraneous x = 11, solution is extraneous

23. anonymous

and none of those are -1 so i don't know what to do next

24. anonymous

@imqwerty @peachpi

25. imqwerty

wait

26. imqwerty

we got x-5 = 1 add 5 to both sides u get x=6

27. anonymous

ok thank you so much

28. anonymous

how do i know if it is extraneous or not

29. imqwerty

and the solution is not extraneous cause when u put x=6 is a true solution :)

30. anonymous

ok so if its a true solution its always not extraneous?

31. imqwerty

yes :)

32. anonymous

ok, thank you for all of your help! it was much appreciated

33. imqwerty

no prblem :)