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Math Help - Need PreCalculus Help

  1. #1
    mkfadeaway11
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    Need PreCalculus Help

    Find all the real solutions of the equation.

    square root of the quantity (x+7), minus the square root of the quantity (x+2)
    equals the square root of the quantity (x-1), minus the square root of the
    quantity (x-2).

    Thanks for your help.
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by mkfadeaway11
    Find all the real solutions of the equation.

    square root of the quantity (x+7), minus the square root of the quantity (x+2)
    equals the square root of the quantity (x-1), minus the square root of the
    quantity (x-2).

    Thanks for your help.
    Hello,

    if you want to eliminate the square roots, you have to square both sides of your equation. Make sure you use the binomial formula correctly and be aware that the transformation of the equation does not give an equivalent equation:

     \sqrt{x+7} - \sqrt{x+2}=\sqrt{x-1}-\sqrt{x-2}
    LHS of equation:
    x+7-2 \cdot \sqrt{x+7} \cdot \sqrt{x+2} + x+2
    RHS of equation:
    x-1-2 \cdot \sqrt{x-1} \cdot \sqrt{x-2}+x-2

    12=2 \sqrt{(x+7)(x+2)}-2\sqrt{(x-1)(x-2)}
    Square both sides again:
    LHS of equation: 144
    RHS of equation:
    4(x+7)(x+2)-8 \sqrt{x^4+6x^3-11x^2-24x+28}+4(x-1)(x-2)
    Isolate the square root 8\sqrt{x^4+6x^3-11x^2-24x+28}on one side of your equation:
    8 \cdot \sqrt{...}=8x^2+24x+64
    Divide first by 8 and then square both sides of your equation. Use (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc with your RHS!
    After a whole bunch of transformation you'll get:
    36x^2+72x+36=0
    36 \cdot (x+1)^2=0
    To solve this equation is certainly no problem for you.
    You have to prove whether your solution fits into the original equation or not, because you have made often no equavelant transformation!

    Bye
    Last edited by CaptainBlack; February 1st 2006 at 12:41 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by earboth
    :
    :
    <br />
\sqrt{x+7} - \sqrt{x+2}=\sqrt{x-1}-\sqrt{x-2}<br />
    :
    [snip]
    :
    36 \cdot (x+1)^2=0
    To solve this equation is certainly no problem for you.
    You have to prove whether your solution fits into the original equation or not, because you have made often no equavelant transformation!

    Bye
    Can't be right as this has root x=-1 which leaves the RHS of
    original equation imaginary and the LHS real.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by mkfadeaway11
    Find all the real solutions of the equation.

    square root of the quantity (x+7), minus the square root of the quantity (x+2)
    equals the square root of the quantity (x-1), minus the square root of the
    quantity (x-2).

    Thanks for your help.
    Find all real solutions of:

    \sqrt{x+7}-\sqrt{x+2}=\sqrt{x-1}-\sqrt{x-2}.

    Squaring both sides and simplifying:

    6-\sqrt{x-7} \sqrt(x+2)=-\sqrt{x-1} \sqrt{x-2}.

    Squaring again and simplifying:

    \sqrt{x+7}\sqrt{x+2}=4+x.

    Squaring again:

    (x+7)(x+2)=(4+x)^2,

    expanding and simplifying give:

    x=2.

    Substituting this back into the original equation confirms that this is
    indeed a solution, and as we have not lost any solutions in our manipulations
    we have that it is the only solution.

    RonL
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  5. #5
    Super Member
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    Quote Originally Posted by CaptainBlack
    Can't be right as this has root x=-1 which leaves the RHS of
    original equation imaginary and the LHS real.

    RonL
    Hello,

    you're right. I've forgotten the 144 on the LHS of the equation. I'm awfully sorry!

    This time: red hot ears.

    Bye
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