anonymous
  • anonymous
solving a radical equation
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
2x+15=5 2x=-10 x=-5, just square on both sides
anonymous
  • anonymous
what about this one?
1 Attachment

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HELP!!!!
  • HELP!!!!
-5
anonymous
  • anonymous
which one were you referring too? @HELP!!!!
HELP!!!!
  • HELP!!!!
the first one
Schrodinger
  • Schrodinger
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?
anonymous
  • anonymous
x = -6 this is the answer i came up with
Schrodinger
  • Schrodinger
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?
Schrodinger
  • Schrodinger
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.
anonymous
  • anonymous
so x=45?
Schrodinger
  • Schrodinger
Yes!
anonymous
  • anonymous
thanks :D
Schrodinger
  • Schrodinger
Np dude.
mathstudent55
  • mathstudent55
|dw:1357102459337:dw|
Schrodinger
  • Schrodinger
Oh, crap. You're right. I totally missed that.
anonymous
  • anonymous
thankyou
mathstudent55
  • mathstudent55
Np, I'm glad I caught it.
Schrodinger
  • Schrodinger
Pardon me about that, @itsjustme_lol.
mathstudent55
  • mathstudent55
Also, whenever you square both sides of an equation, you must check every solution bec squaring both sides can introduce extraneous solutions.
anonymous
  • anonymous
its ok! at least you know that you made a mistke
HELP!!!!
  • HELP!!!!
lol thats why it "open" study

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