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solving a radical equation

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1 Attachment
2x+15=5 2x=-10 x=-5, just square on both sides
what about this one?
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Other answers:

which one were you referring too? @HELP!!!!
the first one
Hey, @itsjustme_lol. Given your second equation, \[-5\sqrt{x-9} - 7 = 23\]Since you're dealing with integers that can be divided and multiplied by each other and still result in even integers, it doesn't have to get nasty. <*hint* So, do you have any clue as to do next?
x = -6 this is the answer i came up with
Okay. Just to get a general idea of how you think and to make things easier for the future, let's say we have \[2\sqrt{x + 4} - 4 = 8\]What would be the first thing you would do, given a square root times something, and two numbers that can be added together?
Ugh. I don't want to get bogged down, nevermind. Okay, so just as a general idea, if you ever have a square root with multiple terms that's being multiplied something, and you can't just square everything without anything getting "ugly" because of how squaring stuff might affect other numbers, almost always, the best thing for you to do is add like terms. In your problem, it would be adding negative seven to the side with twenty-three. Then you would get thirty. From this point, you have a square root with a coefficient, and all of that equals 30. From there, you can divide by that coefficient, and then square everything. Then you can solve for x.
so x=45?
thanks :D
Np dude.
Oh, crap. You're right. I totally missed that.
Np, I'm glad I caught it.
Pardon me about that, @itsjustme_lol.
Also, whenever you square both sides of an equation, you must check every solution bec squaring both sides can introduce extraneous solutions.
its ok! at least you know that you made a mistke
lol thats why it "open" study

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