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itsmylife
 one year ago
Best ResponseYou've already chosen the best response.12x+15=5 2x=10 x=5, just square on both sides

itsjustme_lol
 one year ago
Best ResponseYou've already chosen the best response.1what about this one?

itsjustme_lol
 one year ago
Best ResponseYou've already chosen the best response.1which one were you referring too? @HELP!!!!

Schrodinger
 one year 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?

itsjustme_lol
 one year ago
Best ResponseYou've already chosen the best response.1x = 6 this is the answer i came up with

Schrodinger
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1357102459337:dw

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

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

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

mathstudent55
 one year 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.

itsjustme_lol
 one year ago
Best ResponseYou've already chosen the best response.1its ok! at least you know that you made a mistke

itsjustme_lol
 one year ago
Best ResponseYou've already chosen the best response.1http://openstudy.com/study#/updates/50e3b381e4b028291d74c423

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