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anonymous
 4 years ago
solving a radical equation
anonymous
 4 years ago
solving a radical equation

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02x+15=5 2x=10 x=5, just square on both sides

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which one were you referring too? @HELP!!!!

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0Hey, @itsjustme_lol. Given your second equation, \[5\sqrt{x9}  7 = 23\]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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x = 6 this is the answer i came up with

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0Okay. 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?

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0Ugh. 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 twentythree. 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.

mathstudent55
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1357102459337:dw

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, crap. You're right. I totally missed that.

mathstudent55
 4 years ago
Best ResponseYou've already chosen the best response.1Np, I'm glad I caught it.

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0Pardon me about that, @itsjustme_lol.

mathstudent55
 4 years ago
Best ResponseYou've already chosen the best response.1Also, whenever you square both sides of an equation, you must check every solution bec squaring both sides can introduce extraneous solutions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its ok! at least you know that you made a mistke

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/50e3b381e4b028291d74c423

HELP!!!!
 4 years ago
Best ResponseYou've already chosen the best response.0lol thats why it "open" study
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