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Math Help - Radical-Radical equation

  1. #1
    Super Member dhiab's Avatar
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    Radical-Radical equation

    Solve in R this equation :

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  2. #2
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    Quote Originally Posted by dhiab View Post
    Solve in R this equation :

    1. Substitute u = \sqrt{x-1}

    then u^2 = x-1

    2. \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1 becomes:

    \sqrt{u^2+4-4u}+\sqrt{u^2+9-6u}=1 which is:

    \sqrt{(u-2)^2}+\sqrt{(u-3)^2}=1 and you get:

    |u-2|+|u-3|=1~\implies~2\leq u \leq 3

    3. Now resubstitute:

    2\leq \sqrt{x-1} \leq 3

    and solve for x

    4. This part is for you.
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  3. #3
    Super Member dhiab's Avatar
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    Hello : Thank you , but I'have one question :
    why condition? 2=<u=<3
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  4. #4
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    Quote Originally Posted by dhiab View Post
    Hello : Thank you , but I'have one question :
    why condition? 2=<u=<3
    That isn't a condition but the solution of the equation!

    Use the definition of the absolute value: |a|=\left\{\begin{array}{r}a, a \geq 0\\-a, a < 0\end{array}\right.

    1. |u-2|= \left\{\begin{array}{r}u-2, u \geq 2\\-(u-2), u < 2\end{array}\right.

    2. |u-3|= \left\{\begin{array}{r}u-3, u \geq 3\\-(u-3), u < 3\end{array}\right.

    3. You get three different domains of the equation:

    u\geq 3 : u-2+u-3=1~\implies~2u=6~\implies~u=3

    2 \leq  u < 3 : u-2-u+3=1~\implies~0=0~\implies~2 \leq  u < 3

    u < 2 : -u+2-u+3=1~\implies~-2u=-4~\implies~u=2

    Now collect all partial solutions and you'll get: 2 \leq  u \leq 3
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  5. #5
    Super Member dhiab's Avatar
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    Quote Originally Posted by earboth View Post
    That isn't a condition but the solution of the equation!

    Use the definition of the absolute value: |a|=\left\{\begin{array}{r}a, a \geq 0\\-a, a < 0\end{array}\right.

    1. |u-2|= \left\{\begin{array}{r}u-2, u \geq 2\\-(u-2), u < 2\end{array}\right.

    2. |u-3|= \left\{\begin{array}{r}u-3, u \geq 3\\-(u-3), u < 3\end{array}\right.

    3. You get three different domains of the equation:

    u\geq 3 : u-2+u-3=1~\implies~2u=6~\implies~u=3

    2 \leq u < 3 : u-2-u+3=1~\implies~0=0~\implies~2 \leq u < 3

    u < 2 : -u+2-u+3=1~\implies~-2u=-4~\implies~u=2

    Now collect all partial solutions and you'll get: 2 \leq u \leq 3
    Hello, THANK YOU for resolution
    Now I'have :



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  6. #6
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    Quote Originally Posted by dhiab View Post
    Hello, THANK YOU for resolution
    Now I'have :



    I'm not sure what exactly you have done ...

    As I wrote in my previous post:

    3. Now resubstitute:

    2\leq \sqrt{x-1} \leq 3

    and solve for x
    you only have to get rid of the square-root. So square the complete inequality and you'll get:

    2\leq \sqrt{x-1} \leq 3~\implies~4\leq x-1 \leq 9~\implies~\boxed{5\leq x \leq 10}

    I've attached the graph of the function.
    Attached Thumbnails Attached Thumbnails Radical-Radical equation-wurzwurzglg.png  
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  7. #7
    Super Member dhiab's Avatar
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    Quote Originally Posted by earboth View Post
    I'm not sure what exactly you have done ...

    As I wrote in my previous post:



    you only have to get rid of the square-root. So square the complete inequality and you'll get:

    2\leq \sqrt{x-1} \leq 3~\implies~4\leq x-1 \leq 9~\implies~\boxed{5\leq x \leq 10}

    I've attached the graph of the function.
    THANK YOU
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