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

which expression is equivalent to x?
\[b^{2}-a \]

\[b-a\]

b+a

or b^2+a

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

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

It will just get rid of it :)!!

so b^2+ a is the answer?

yes

how did you know x=> ?

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

no i dont

|dw:1358391793350:dw|

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

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

where did you get the square
from?

ohh ok

thanks so much :)

No problem :)

can you help me with another problem?

SUre

yea?

i thought you left

I can help your with your question if you need

you*