How to Solve 4\sqrt{x-2} + 8=3\sqrt{x+6}-5 for x | Refresh Your Memory

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To solve the equation -4√(x-2) + 8 = 3√(x+6) - 5 for x, begin by isolating the square roots and then square both sides to eliminate them. This process may need to be repeated to fully solve for x. Care must be taken to ensure that the signs of both sides are the same when squaring, and that no negative values are introduced under the square roots. After finding potential solutions, verify them against the original equation, as some may need to be discarded. Understanding these steps is essential for successfully solving similar irrational equations.
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my teacher showed me how to solve this but she said we don't have to know it till cegep. id still like to know how to do it, can u guys help me refresh my memory?

-4\sqrt{x-2} + 8=3\sqrt{x+6}-5

solve for x
 
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You can solve these kind of irrational equations by squaring both sides, it may be necessary to do so more than once.
You square because you want to get rid off the square roots, so it's necessary to carefully choose what to put on what side because you don't want to create new square roots as a result of mixed terms, or at least you want to minimize the number of those.

You have to be careful though, when squaring - the sign of both sides has to be the same and all expressions under the square roots can't be negative, these constraints give you conditions to test on your final answers - some may have to be cancelled.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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