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Math Help - Find a radical equation that has solution x=5 and extraneous solutions x=-2 and x=-1

  1. #1
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    Find a radical equation that has solution x=5 and extraneous solutions x=-2 and x=-1

    I found a radical equation, the equation is
    (Square root of(2x^3-2x+16))= (2x+6).

    When I check my equation to make sure the above equation generates
    extraneous solution x=-1 and x=-2. I get all the three x=5, x=-1 and
    x=-2 as solutions, all three satisfy my equation.

    Can anybody comment on where I am going wrong
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  2. #2
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    Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and

    How did you find it?

    Could you start by expanding (x-5)(x+2)(x+1) ?
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    Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and

    Yes, I started by expanding (x-5)(x+2)(x+1)
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    Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and

    Then what did you do?
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    Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and

    Look at the attached document
    [IMG]file:///C:/Users/MEYYAP%7E1/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
    Radical.doc
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  6. #6
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    Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and

    Quote Originally Posted by anna123 View Post
    Yes, I started by expanding (x-5)(x+2)(x+1)
    Expand that & set it equal to zero, then subtract a perfect square which is positive at x = -1 & x = -2, but is zero at x = 5. The simplest one of those is (x-5)^2.

    Then take the square root of both sides of the resulting equation.
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  7. #7
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    Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and

    After expanding, I am getting x^3-2x^2-13x-10=0. Are you saying I have to subtract (x-5)^2 from both sides?

    x^3-2x^2-13x-10-(x-5)^2 = 0 - (x-5)^2?????
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  8. #8
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    Re: Find an radical equation that has solution x=5 and extraneous solution x=-2 and

    Yep, now expand -(x-5)^2 on the LHS and group like terms, then take the square root of both sides, then you are finished.
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    Re: Find a radical equation that has solution x=5 and extraneous solutions x=-2 and

    Thank you for your help!

    I did subtract (x-5) from both sides and I ended up with the equation [IMG]file:///C:/Users/MEYYAP%7E1/AppData/Local/Temp/moz-screenshot.png[/IMG](x-5)= square root of (3x^2+3x+35-x^3).

    Can you explain your reasoning behind why you wanted me to subtract x-5 from both sides?
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  10. #10
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    Re: Find a radical equation that has solution x=5 and extraneous solutions x=-2 and

    Quote Originally Posted by anna123 View Post
    Thank you for your help!

    I did subtract (x-5) from both sides and I ended up with the equation [IMG]file:///C:/Users/MEYYAP%7E1/AppData/Local/Temp/moz-screenshot.png[/IMG](x-5)= square root of (3x^2+3x+35-x^3).

    Can you explain your reasoning behind why you wanted me to subtract x-5 from both sides?
    Oh fiddlesticks !!! Now I see the problem with that!!! -(x-5)^2 isn't a perfect square.

    Starting from scratch:

    I like to take something like (\sqrt{x+2}-5)(\sqrt{x+2}+3)=0\,, then multiply this out. The resulting equation will be solved by isolating the radical & squaring both sides. But if you look the above equation, the second factor, (\sqrt{x+2}+3) can not be zero, because the principal square root is not negative. But if you set this factor to zero, subtract 2 then square it, you get x+2 = 9 → x = 7, which will be an extraneous solution, if the following equation is solved conventionally. The other factor gives x = 23.

    Multiplying the above equation out, gives:

    x-2\sqrt{x+2}=13 or any equivalent equation.

    For a solution x = 5, and an extraneous solution of x = -2, try: (\sqrt{x+11}-4)(\sqrt{x+11}+3)=0\,.

    However, I find it very difficult to get one valid solution AND two extraneous solutions.
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  11. #11
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    Re: Find a radical equation that has solution x=5 and extraneous solutions x=-2 and

    I think this problem is easier than you are trying to make it. Expanding (x-5)(x+2)(x+1) as pickslides suggested, you get the equation x^3-2x^2 - 13x - 10=0. You want to "radicalise" this in a way that works for the positive solution x=5 but not for the two negative solutions. So write it as x^2 = \tfrac12(x^3-13x-10). Now take the square root to get \boxed{x = \sqrt{\tfrac12(x^3-13x-10)}}.
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