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## ligia587 Group Title which is expression is equivalent to one year ago one year ago

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1. ligia587 Group Title

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

2. ligia587 Group Title

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

3. ligia587 Group Title

$b-a$

4. ligia587 Group Title

b+a

5. ligia587 Group Title

or b^2+a

6. kirbykirby Group Title

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

7. kirbykirby Group Title

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

8. kirbykirby Group Title

It will just get rid of it :)!!

9. ligia587 Group Title

so b^2+ a is the answer?

10. kirbykirby Group Title

yes

11. ligia587 Group Title

how did you know x=> ?

12. kirbykirby Group Title

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

13. ligia587 Group Title

no i dont

14. ligia587 Group Title

|dw:1358391793350:dw|

15. kirbykirby Group Title

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

16. kirbykirby Group Title

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 Group Title

where did you get the square from?

18. kirbykirby Group Title

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 Group Title

ohh ok

20. ligia587 Group Title

thanks so much :)

21. kirbykirby Group Title

No problem :)

22. ligia587 Group Title

can you help me with another problem?

23. kirbykirby Group Title

SUre

24. kirbykirby Group Title

@ligia587

25. ligia587 Group Title

yea?

26. ligia587 Group Title

i thought you left

27. kirbykirby Group Title

I can help your with your question if you need

28. kirbykirby Group Title

you*