## airyana1114 one year ago Can someone help me solve for x?

1. airyana1114

$2\sqrt{x-5}=2$

2. airyana1114

I know my first step is to subtract 2 from both sides of the equation which gives me... $\sqrt{x-5}=0$ Then I subtract 5 from both sides and I'm left with... $\sqrt{x}=5$ Is there another step after that?

3. zepdrix

Wooooops! :O

4. zepdrix

If the 2 was being added to the root, then yes, you would subtract it from both sides to undo the addition.

5. zepdrix

But in this case, it looks like the 2 is multiplying the root, yes?

6. zepdrix

So how do we undo multiplication? :)

7. airyana1114

We divide :D

8. zepdrix

Good, yes :) Do that. It should give you something a little different than 0 on the right side.

9. airyana1114

Okay, so now I got 1 instead of 0. @zepdrix

10. zepdrix

$\large\rm \sqrt{x-5}=1$Cool.

11. airyana1114

Then I would add 5 and get $\sqrt{x}=6$ @zepdrix

12. zepdrix

Wooops! The subtraction is inside of the root. We can't do that operation yet. We need to deal with the root first.

13. zepdrix

What's the inverse of square root`? How do we undo that? :o

14. airyana1114

Okay so square both sides of the equation? It would still be 1 correct?

15. zepdrix

$\large\rm \left(\sqrt{x-5}\right)^2=1^2$On the left, the operations undo one another, good. On the right, 1^2 = 1? Sounds good!$\large\rm x-5=1$

16. airyana1114

Okay, then I would add 5 to both sides of the equation correct? @zepdrix

17. anonymous

yes

18. airyana1114

Thank you :) @satellite73

19. zepdrix

yay good job \c:/