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ligia587
 one year ago
Best ResponseYou've already chosen the best response.0if \[\sqrt{xa}=b, x>a \]

ligia587
 one year ago
Best ResponseYou've already chosen the best response.0which expression is equivalent to x? \[b^{2}a \]

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

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

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

ligia587
 one year ago
Best ResponseYou've already chosen the best response.0so b^2+ a is the answer?

ligia587
 one year ago
Best ResponseYou've already chosen the best response.0how did you know x=> ?

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

ligia587
 one year ago
Best ResponseYou've already chosen the best response.0dw:1358391793350:dw

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

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

ligia587
 one year ago
Best ResponseYou've already chosen the best response.0where did you get the square from?

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.1It'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

ligia587
 one year ago
Best ResponseYou've already chosen the best response.0can you help me with another problem?

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