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How can I move one term with Sqrt to the right side of equation?

The root of the previous problem does not have a coefficient in front of it, so using the previous code for this problem has no effect.

5 (-5 + Sqrt[x^2 + (-2 + y)^2]) + 2 y == 0

How to identify a radical with coefficients in front of it and move it to the right of the equation

to get the result:

-25 + 2 y == -5 Sqrt[x^2 + (-2 + y)^2]
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1 Answer 1

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You should put more effort in understanding the answers you've obtained:

Expand[5 (-5 + Sqrt[x^2 + (-2 + y)^2]) + 2 y == 0] /. 
 a_ + b_ Sqrt[c_] == 0 :> a == -b Sqrt[c]
(* -25 + 2 y == -5 Sqrt[x^2 + (-2 + y)^2] *)
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  • $\begingroup$ In[654]:= Expand[5 (-5 + Sqrt[x^2 + (-2 + y)^2]) + 2 y == 0] /. a_ + b_.*Sqrt[c_] == 0 :> a == -b Sqrt[c] Out[654]= -25 + 2 y == -5 Sqrt[x^2 + (-2 + y)^2] In[655]:= Expand[(-5 + Sqrt[x^2 + (-2 + y)^2]) + 2 y == 0] /. a_ + b_.*Sqrt[c_] == 0 :> a == -b Sqrt[c] Out[655]= -5 + 2 y == -Sqrt[x^2 + (-2 + y)^2] $\endgroup$
    – csn899
    Commented Aug 16, 2023 at 1:12

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