## ligia587 2 years ago which is expression is equivalent to

1. ligia587

if $\sqrt{x-a}=b, x>a$

2. ligia587

which expression is equivalent to x? $b^{2}-a$

3. ligia587

$b-a$

4. ligia587

b+a

5. ligia587

or b^2+a

6. kirbykirby

$\sqrt{x-a}=b$=>$x-a=b^2$=>$x=b^2+a$

7. kirbykirby

When solving with radicals, square both sides because $(\sqrt{x})^2=x$

8. kirbykirby

It will just get rid of it :)!!

9. ligia587

so b^2+ a is the answer?

10. kirbykirby

yes

11. ligia587

how did you know x=> ?

12. kirbykirby

Did you understand the first step? that x-a = b^2 ?

13. ligia587

no i dont

14. ligia587

|dw:1358391793350:dw|

15. kirbykirby

Ok. Do you understand why $(\sqrt{x})^2=x$?

16. kirbykirby

The "x > a" is just imposing a restriction value on your x. You don't need to use it to solve for x

17. ligia587

where did you get the square from?

18. kirbykirby

It's what you need to do in order to solve the equation. It's similar to like when you have an equation like: $2x-4=10$$2x=10+4$$2x=14$$\frac{2x}{2}=\frac{14}{2}$$x=7$ You could argue "how did you get the divided by 2 part?" Well it's similar to your situation. You introduce this exponent to get rid of your square root, just like you introduced a divided by 2 up here^ to isolate your x

19. ligia587

ohh ok

20. ligia587

thanks so much :)

21. kirbykirby

No problem :)

22. ligia587

can you help me with another problem?

23. kirbykirby

SUre

24. kirbykirby

@ligia587

25. ligia587

yea?

26. ligia587

i thought you left

27. kirbykirby