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

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

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

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

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

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

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

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

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

kirbykirby
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0where did you get the square from?

kirbykirby
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0can you help me with another problem?

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