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ligia587Best ResponseYou've already chosen the best response.0
if \[\sqrt{xa}=b, x>a \]
 one year ago

ligia587Best ResponseYou've already chosen the best response.0
which expression is equivalent to x? \[b^{2}a \]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
\[\sqrt{xa}=b\]=>\[xa=b^2\]=>\[x=b^2+a\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
When solving with radicals, square both sides because \[(\sqrt{x})^2=x\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
It will just get rid of it :)!!
 one year ago

ligia587Best ResponseYou've already chosen the best response.0
so b^2+ a is the answer?
 one year ago

ligia587Best ResponseYou've already chosen the best response.0
how did you know x=> ?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
Did you understand the first step? that xa = b^2 ?
 one year ago

ligia587Best ResponseYou've already chosen the best response.0
dw:1358391793350:dw
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
Ok. Do you understand why \[(\sqrt{x})^2=x\]?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
The "x > a" is just imposing a restriction value on your x. You don't need to use it to solve for x
 one year ago

ligia587Best ResponseYou've already chosen the best response.0
where did you get the square from?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
It's what you need to do in order to solve the equation. It's similar to like when you have an equation like: \[2x4=10\]\[2x=10+4\]\[2x=14\]\[\frac{2x}{2}=\frac{14}{2}\]\[x=7\] 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
 one year ago

ligia587Best ResponseYou've already chosen the best response.0
can you help me with another problem?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
I can help your with your question if you need
 one year ago
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