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Honors pre-calculus homework, help

  1. Oct 26, 2011 #1
    honors pre-calculus homework, help!!

    1. The problem statement, all variables and given/known data
    find the sum of the roots of the equation (x-1)^1/2 + (2x-1)^1/2 = x


    2. Relevant equations
    I have no idea, I just started a pre calculus course about 5 weeks ago and our teacher gave the people in honors problems we've never seen before so i'm not sure what would be a relevant equation


    3. The attempt at a solution
    I know I need to put the problem in standard form before i can find the roots, but would that look like this:(x-1)^1/2 + (2x-1)^1/2 -x = y ? I'm not sure exactly how to start this one...
     
  2. jcsd
  3. Oct 26, 2011 #2
    Re: honors pre-calculus homework, help!!

    How might you get rid of the square roots?
     
  4. Oct 27, 2011 #3

    symbolipoint

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    Re: honors pre-calculus homework, help!!

    To find the roots, you first should use algebraic properties to transform the given equation into a polynomial equation of degree 2 (meaning quadratic equation; something times something squared plus something else times the something plus a constant equals zero). Knowing how would be a result of your intermediate algebra. Once you find the roots, the solutions to the original equation, just finish by adding them, according to the given instructions, "find the sum of the roots.."
     
  5. Oct 27, 2011 #4
    Re: honors pre-calculus homework, help!!

    ok thanks both of you!
     
  6. Oct 29, 2011 #5

    NascentOxygen

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    Re: honors pre-calculus homework, help!!

    What steps did you go through to solve this, hpdwnsn95?
     
  7. Oct 29, 2011 #6

    symbolipoint

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    Re: honors pre-calculus homework, help!!

    If he said he solved it, then I believe it. I also solved it myself. Answer was "2". Not difficult. One false root occurred in the solution process, eliminated by division of both members of the equation. Two "roots" remaining, easily found and processed.
     
  8. Oct 29, 2011 #7

    LCKurtz

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    Re: honors pre-calculus homework, help!!

    Interesting that you got 2 for the sum of the roots when they are x = 1 and x = 5.
     
  9. Oct 29, 2011 #8

    NascentOxygen

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    Re: honors pre-calculus homework, help!!

    I had difficulty understanding the suggested method/s of solving this, and hoped hpdwnsn95's answer would clear up what I was missing. But thank you for butting in just the same.
     
  10. Oct 29, 2011 #9

    LCKurtz

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    Re: honors pre-calculus homework, help!!

    Me too. I was thinking the problem might be suggesting some clever way to find the sum of the roots without brute force squaring it out. Once you have the roots it seems pretty silly to ask what the sum of them is.
     
  11. Oct 29, 2011 #10

    symbolipoint

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    Re: honors pre-calculus homework, help!!

    With that analysis, my own work deserves more careful checking. My fourth attempt seems to show this set of steps to reach the quadratic equation stage of the solution:

    [tex]\[
    \begin{array}{l}
    (x - 1)^{1/2} + (2x - 1)^{1/2} = x \\
    (x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} = x - (2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} \\
    x - 1 = [x - (2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} ]^2 \\
    x - 1 = x^2 - 2x(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} + 2x - 1 \\
    x - 1 - 2x + 1 = x^2 - 2x(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} \\
    - x = x^2 - 2x(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} \\
    - x - x^2 = - 2x(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} \\
    x + 1 = 2(2x - 1)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} \\
    x^2 + 2x + 1 = 4(2x - 1) \\
    x^2 + 2x + 1 = 8x - 4 \\
    x^2 - 6x + 5 = 0 \\
    \end{array}
    \]
    [/tex]

    Remaining to be done is find the actual roots and then their sum.
     
  12. Oct 30, 2011 #11

    NascentOxygen

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    Re: honors pre-calculus homework, help!!

    Squaring it out twice, at that.

    But once you have x2 - 6x + 5
    remember...

    -(sum of roots) = -6, and
    product of roots = +5

    Though, given this is a pre-calculus exercise, I think the recommended first option should have been to try the whole numbers from, say, -10 to +10. We can scratch all the negatives, since RHS must be positive as it results from addition, then one at a time try from 1..upwards.
     
  13. Oct 30, 2011 #12

    LCKurtz

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    Re: honors pre-calculus homework, help!!

    Yes, I knew that. But if you have to go through the work required to get it to there, finding the roots is so trivial that being asked to find the sum of the roots is pretty much pointless. Now if there were some clever way of getting the sum of the roots from the original equation, without finding the roots, that would would make it an interesting problem. And since you are asked for the sum of the roots and not the roots themselves, I wondered if there wasn't some clever trick that I haven't seen before.
     
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