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Math Help - Solve this radical equation

  1. #1
    Super Member dhiab's Avatar
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    Solve this radical equation

    Solve in R :
    \sqrt {x - 1} + \sqrt {2 - x} = x^2 - 3x + 3
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  2. #2
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    Square both sides. You'll end up with a radical as the middle term on the right-hand side.

    Isolate this radical, and then square again.

    Then solve the resulting polynomial.

    Remember to check your solutions in the original radical equation!

    If you get stuck, please reply showing your steps and reasoning so far. Thank you!
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by stapel View Post
    Square both sides. You'll end up with a radical as the middle term on the right-hand side.

    Isolate this radical, and then square again.

    Then solve the resulting polynomial.

    Remember to check your solutions in the original radical equation!

    If you get stuck, please reply showing your steps and reasoning so far. Thank you!
    Hello :I'have
    \begin{array}{l}<br />
\forall x \in \Re :x^2 - 3x + 3 > 0 \\ <br />
because:\Delta = - 3 < 0 \\ <br />
then: \\ <br />
x - 1 + 2 - x + 2\sqrt {\left( {x - 1} \right)\left( {2 - x} \right)} = \left( {x^2 - 3x + 3} \right)^2 \\ <br />
2\sqrt {\left( {x - 1} \right)\left( {2 - x} \right)} = \left( {x^2 - 3x + 3} \right)^2 - 1??????? \\ <br />
\end{array}
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  4. #4
    MHF Contributor red_dog's Avatar
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    \sqrt{x-1}+\sqrt{2-x}=-(x-1)(2-x)+1

    Let a=\sqrt{x-1}, \ b=\sqrt{2-x}

    Then we have the system of equations:

    \left\{\begin{array}{ll}a+b=-a^2b^2+1\\a^2+b^2=1\end{array}\right.

    Let S=a+b, \ P=ab. Then

    \left\{\begin{array}{ll}S+P^2=1\\S^2-2P=1\end{array}\right.

    From the first equation S=1-P^2

    Replace in the second:

    P(P^3-2P-2)=0

    If P=0 then S=1.

    Then \left\{\begin{array}{ll}a+b=1\\ab=0\end{array}\rig  ht.

    We get \left\{\begin{array}{ll}a=1\\b=0\end{array}\right.  \Rightarrow x=2

    \left\{\begin{array}{ll}a=0\\b=1\end{array}\right.  \Rightarrow x=1

    Now we have to see what happens if P^3-2P^2-2=0

    But this equation has no rational roots and the discussion is more complicated.
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  5. #5
    Senior Member pacman's Avatar
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    from red_dog, p^3 - 2p^2 - 0 = 0, by iteration; p > 2.359.

    the graph confirms it
    Attached Thumbnails Attached Thumbnails Solve this radical equation-pit.gif  
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  6. #6
    Super Member dhiab's Avatar
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    Quote Originally Posted by red_dog View Post
    \sqrt{x-1}+\sqrt{2-x}=-(x-1)(2-x)+1

    Let a=\sqrt{x-1}, \ b=\sqrt{2-x}

    Then we have the system of equations:

    \left\{\begin{array}{ll}a+b=-a^2b^2+1\\a^2+b^2=1\end{array}\right.

    Let S=a+b, \ P=ab. Then

    \left\{\begin{array}{ll}S+P^2=1\\S^2-2P=1\end{array}\right.

    From the first equation S=1-P^2

    Replace in the second:

    P(P^3-2P-2)=0

    If P=0 then S=1.

    Then \left\{\begin{array}{ll}a+b=1\\ab=0\end{array}\rig  ht.

    We get \left\{\begin{array}{ll}a=1\\b=0\end{array}\right.  \Rightarrow x=2

    \left\{\begin{array}{ll}a=0\\b=1\end{array}\right.  \Rightarrow x=1

    Now we have to see what happens if P^3-2P^2-2=0

    But this equation has no rational roots and the discussion is more complicated.
    HELLO :
    This equation p^3 - 2p^2 - 2 = 0<br />
has not real roots because I'have 0 \le p \le 1
    and if

    ^3 - 2p^2 - 2 < 0" alt="p \in \left[ {0,1} \right]^3 - 2p^2 - 2 < 0" />
    There is the details :

    1 \le x \le 2.....because:\left( {x - 1 \ge 0} \right) \wedge \left( {2 - x \ge 0} \right)<br />
    Then :
    \begin{array}{l}<br />
\\ <br />
1 \le x \le 2 \Leftrightarrow 0 \le x - 1 \le 1 \Leftrightarrow 0 \le \sqrt {x - 1} \le 1......(1) \\ <br />
1 \le x \le 2 \Leftrightarrow - 1 \ge - x \ge - 2 \Leftrightarrow 0 \le 2 - x \le 1 \Leftrightarrow 0 \le \sqrt {2 - x} \le 1.....(2) \\ <br />
\end{array}

    but:a = \sqrt {x - 1} ...b = \sqrt {x - 1}
    by (1),(2) :

    0 \le p \le 1<br />
    THANKS
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