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Math Help - Principle of Powers problem

  1. #1
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    Principle of Powers problem

    Hi, first post:
    I'm helping a student work a radical equation and have a question. Here is the problem:

    √2X-5 = 1 + √X - 3 (2X - 5 AND X - 3 ARE ALL UNDER THE RADICAL SIGN)

    Step 2: (√2x 5)2 = (1 + √x-3)2

    Step 3: 2x 5 = 1 = 2√x-3 + (x -3 ) My question is HOW did they get the 2 in front of the radical sign!!!

    Step 4: x 3 = 2√x-3 so Im really lost on this one.

    The answer is x = 7 or x = 3 but WHAT did they do between step 2 and step 3 to get the 2 in front of the radical sign?

    Thanks.


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  2. #2
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    Hello,

    Quote Originally Posted by lepraconn View Post
    Hi, first post:
    I'm helping a student work a radical equation and have a question. Here is the problem:

    √2X-5 = 1 + √X - 3 (2X - 5 AND X - 3 ARE ALL UNDER THE RADICAL SIGN)

    Step 2: (√2x – 5)2 = (1 + √x-3)2

    Step 3: 2x – 5 = 1 = 2√x-3 + (x -3 ) My question is HOW did they get the 2 in front of the radical sign!!!


    ~~
    \left(1+\sqrt{x-3}\right)^2=(a+b)^2=a^2+{\color{red}2}ab+b^2

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  3. #3
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    Quote Originally Posted by lepraconn View Post
    Hi, first post:
    I'm helping a student work a radical equation and have a question. Here is the problem:

    √(2X-5) = 1 + √(X - 3)
    (2X - 5 AND X - 3 ARE ALL UNDER THE RADICAL SIGN)
    ...
    I'll do this question step by step so you have an example how to do the next 166 questions of the same type:

    1. In general: To get rid of the square-root you have to square - but correctly:

    (a * b) = a * b

    (a+b) = a + 2ab + b

    (a-b) = a - 2ab + b

    2. The tactics in solving such an equation is: Square - isolate remainings roots - square - isolate .... until there isn't any root left:

    3. Your question:

    \sqrt{2x-5} = 1+\sqrt{x-3}~,~x \geq 3 ........ square both sides

    2x-5=1+2 \cdot 1 \cdot \sqrt{x-3} + x-3 ........ collect like terms and isolate the root on the RHS of the equation

    x-3 = 2\sqrt{x-3} ........ square both sides
    x^2-6x+9=4(x-3) ........ collect like terms

    x^2-10x+21=0........ use the formula to solve this quadratic equation. I've got:

    x = 3 ~\vee~ x= 7

    Since squaring is not an equivalent transformation you have to plug in the results into the original equation and check if the solutions satisfy the equation:

    1. x = 3~\implies~ \sqrt{2 \cdot 3-5} = 1+\sqrt{3-3} ~\implies~ \sqrt{1} = 1

    2. x = 7 ~\implies~ \sqrt{2 \cdot 7-5} = 1+\sqrt{7-3} ~\implies~ \sqrt{9} = 1 + \sqrt{4}

    Obviously both results are solutions of the equation.
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  4. #4
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    additional remark

    I've demonstrated the more usual way to solve such a type of questions. With your question you may use a short cut at this point:

    x-3 = 2\sqrt{x-3}

    Use the substitution: \sqrt{x-3} =y

    Then the equation becomes:

    y^2 = 2y~\implies~y(y-2)=0~\implies~y=0~\vee~y=2

    That means:

    \sqrt{x-3}=0~\implies~x=3

    \sqrt{x-3}=2~\implies~x-3=4~\implies~x=7
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  5. #5
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    Hello, lepraconn!

    Welcome abarod!


    \sqrt{2x-5} \:= \:1 + \sqrt{x-3}

    Step 2: . \left(\sqrt{2x - 5}\right)^2 \:= \:\left(1 + \sqrt{x-3}\right)^2

    Step 3: . 2x - 5 \:= \:1 + 2\sqrt{x-3} + (x-3)


    My question is HOW did they get the 2 in front of the radical sign?

    I don't suppose you even tried to square it out, did you?

    \left(1 + \sqrt{x-3}\right)\left(1 + \sqrt{x-3}\right) \;=\;1^2 + \sqrt{x-3} + \sqrt{x-3} + \left(\sqrt{x-3}\right)^2

    . . Got it?

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