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anonymous
 5 years ago
find a radical equation of the form sqrt(ax+b)=x+c so that one solution is extraneous
anonymous
 5 years ago
find a radical equation of the form sqrt(ax+b)=x+c so that one solution is extraneous

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay. give an example of square root of a number.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so 4 is the square root of 16. so ax+b =16 and x+c = 4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0say x is 3, then c is 1. and 3a+b is 16. lets say a is 4, then b is also 4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so your equation could be sqrt(4x+4)=x+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to find x, square both sides. and post what you got.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait. what is (x+1)^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(a+b)^2 is a^2+b^2? are you sure about that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know the formula for \[(a+b)^{2} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, if you don't know the formula, you can tell me. I will explain.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait i no this its 4x+4=x^2+2x+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes! solve that expression then.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y, plug in x = 3 and x = 1 in your original expression, sqrt(4x+4)=x+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait. this is a bad example.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, anyway, you get the idea. you have to follow the procedure I outlines, such that the original expression is solve by one value of x and the square expression is solved by 2 values of x, giving you one spurious value of x.
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