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Algebra II Equations Containing Radicals Part 2

  1. Jun 19, 2012 #1
    I managed to get the other problems right, but this one I've been fiddling with and can't seem to get the right answer.

    1. The problem statement, all variables and given/known data
    [tex] \sqrt {3b -2} - \sqrt {2b + 5} = 1 [/tex]

    Answer: [itex] 22 [/itex]


    2. Relevant equations

    --


    3. The attempt at a solution

    For this I tried the method of leaving the equation as is and squaring both sides, since this is the way the textbook seems to want me to learn it for this lesson.

    [itex] \sqrt {3b - 2} - \sqrt {2b + 5} = 1 [/itex]
    [itex] (\sqrt {3b - 2} - \sqrt {2b + 5})^2 = 1^2 [/itex]
    [itex] 3b - 2 - 2\sqrt {6b^2 - 10} + 2b + 5 = 1 [/itex]
    [itex] 5b + 3 - 2\sqrt {6b^2 - 10} = 1 [/itex]
    [itex] 5b + 3 = 2\sqrt {6b^2 - 10} [/itex]
    [itex] (5b + 3)^2 = (2\sqrt {6b^2 - 10})^2 [/itex]
    [itex]25b^2 + 30b + 9 = 4(6b^2 - 10) [/itex]
    [itex] 25b^2 + 30b + 9 = 24b^2 - 40 [/itex]
    [itex] b^2 + 30b +49 = 0 [/itex]

    At this point, I realize that 22 isn't a factor of that final equation. I've also tried the other methods, such as [itex] \sqrt {3b - 2} = \sqrt {2b + 5} [/itex], and I get b = 7.

    I've tried to look through it for obvious arithmetic mistakes or my tendency to create incorrect products of the form (a + b)^2, but I didn't see anything. It could be the sheer volume of the problem has me spooked or tripping up.

    Any help would be greatly appreciated. Thanks in advance.
     
  2. jcsd
  3. Jun 19, 2012 #2

    Mark44

    Staff: Mentor

    This is how I would do it, but you have omitted a term.

    The equation should be ##\sqrt {3b - 2} = \sqrt {2b + 5} + 1##.
    Now square both sides.
     
  4. Jun 19, 2012 #3
    Starting from the original equation, move the second radical to the other side and square both sides. You should end up with:

    [itex]3b - 2 = (1 + \sqrt{2b + 5})^2[/itex]

    After foiling, combine like terms and solve for the [itex]2\sqrt{2b + 5}[/itex] term. You should end up with b - 8 on the other side.

    Square both sides once more.

    [itex](b - 8)^2 = 4(2b + 5)[/itex]

    Foil/distribute and set equal to zero.

    Factor the remaining equation and you should end up with zeroes of 2 and 22.
    Plug both answers back into the original equation to see which ones work.

    Good luck :)
     
  5. Jun 19, 2012 #4
    You're missing the +11b term from multiplying out (3b-2)(2b+5).
     
  6. Jun 19, 2012 #5
    You also forgot to subtract the 1 on the right side over to the left side.
     
  7. Jun 20, 2012 #6
    Aaah, I see. Apparently, I have trouble with my factoring instincts. A.k.a. I need to factor things out the old fashioned way as opposed to doing mental math.

    I see where I went wrong and even though my textbook seems to want me to keep the radicals all on one side, I think I'm going to make exceptions for cases such as this. It gets incredibly messy and too advanced for me the other way. Nevertheless, I tried it out just to try and understand what I was missing; and even though, I couldn't arrive at an answer that way, thanks to everyone's pointers I see where I went wrong here again.

    So, thank you Mark44, Hertz (thanks for the luck; I need it! :D), Bohrok, and Villyer.



    Here's what I got the last time:

    [itex] \sqrt {3b -2} - \sqrt {2b + 5} = 1 [/itex]
    [itex] (\sqrt {3b - 2})^2 =(\sqrt {2b + 5}+ 1)^2 [/itex]
    [itex] 3b - 2 = 2b + 5 + 2\sqrt {2b + 5} + 1 [/itex]
    [itex] (b - 8)^2 = (2\sqrt {2b + 5})^2 [/itex]
    [itex] b^2 - 16b + 64 = 4(2b + 5) [/itex]
    [itex] b^2 -16b + 64 = 8b + 20 [/itex]
    [itex] b^2 - 24b + 44 = 0 [/itex]
    [itex] (b - 2)(b - 22) = 0 [/itex]
    b = 2 or b = 22

    '2' doesn't work; it produces 2 - 3 = 1; '22' checks out with an 8 - 7 = 1. So, unless I bent the rules of algebra to solve this, case closed.

    Thanks once again everyone!
     
    Last edited: Jun 20, 2012
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