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Math Help - Algebraic Manipulations and Equations

  1. #1
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    Algebraic Manipulations and Equations

    sqrt(x-27) + sqrt(x) = 9

    solve for x...

    i tried squaring both sides to get the sqrts to go away but i still couldnt get the right answer.

    I need to know how to get to the final solution. FYI the answer is 36.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by johntuan View Post
    sqrt(x-27) + sqrt(x) = 9

    solve for x...

    i tried squaring both sides to get the sqrts to go away but i still couldnt get the right answer.

    I need to know how to get to the final solution. FYI the answer is 36.
    Square it :

    (x-27)+x+2 \sqrt{x(x-27)}=81

    2x+2 \sqrt{x(x-27)}=108

    \sqrt{x(x-27)}=54-x

    Square again and solve the quadratic
    Recall that x-27 \ge 0 and x \ge 0 (which makes x \ge 27)
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  3. #3
    Moo
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    Another way of doing it :

    Let a=\sqrt{x-27} and b=\sqrt{x}

    We have a+b=9

    Now, multiply both sides by (a-b) :
    \underbrace{(a+b)(a-b)}_{a^2-b^2}=9(a-b)

    (x-27)-x=9(a-b)
    -27=9(a-b)
    a-b=-3

    So you have :
    \left\{\begin{array}{ll} a+b=9 \quad (1) \\ a-b=-3 \quad (2) \end{array} \right.

    (1)-(2) ~:~ 2b=12 \implies b=6 \implies \sqrt{x}=6 \implies x=36

    Check if it is correct :
    (1)+(2) ~:~ 2a=6 \implies a=3 \implies \sqrt{x-27}=3 \implies x-27=9 \implies x=36


    Yay !
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  4. #4
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by Moo View Post
    Another way of doing it :

    Let a=\sqrt{x-27} and b=\sqrt{x}

    We have a+b=9

    Now, multiply both sides by (a-b) :
    \underbrace{(a+b)(a-b)}_{a^2-b^2}=9(a-b)

    (x-27)-x=9(a-b)
    -27=9(a-b)
    a-b=-3

    So you have :
    \left\{\begin{array}{ll} a+b=9 \quad (1) \\ a-b=-3 \quad (2) \end{array} \right.

    (1)-(2) ~:~ 2b=12 \implies b=6 \implies \sqrt{x}=6 \implies x=36

    Check if it is correct :
    (1)+(2) ~:~ 2a=6 \implies a=3 \implies \sqrt{x-27}=3 \implies x-27=9 \implies x=36


    Yay !
    I'm probably being dumb once again, but where did those x's go?
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  5. #5
    Moo
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    Quote Originally Posted by shinhidora View Post
    I'm probably being dumb once again, but where did those x's go?
    (x-27)-x=x-27-x=(x-x)-27=-27

    And I kept (a-b) in the left side because we don't need to transform them into x.

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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by Moo View Post
    Hello,

    Square it :

    (x-27)+x+2 \sqrt{x(x-27)}=81

    2x+2 \sqrt{x(x-27)}=108

    \sqrt{x(x-27)}=54-x

    Square again and solve the quadratic
    Recall that x-27 \ge 0 and x \ge 0 (which makes x \ge 27)
    Lookie here! You might like it this way, too. Pardon me, Moo. I just needed to post something. Haven't been on a lot today.

    \sqrt{x-27}+\sqrt{x}=9 Move the \sqrt{x} to the RHS.

    \sqrt{x-27}=9-\sqrt{x} Square both sides.

    x-27=81-18\sqrt{x}+x Collect terms.

    -108=-18\sqrt{x} Square both sides again.

    11664=324x Divide

    \boxed{x=36}
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  7. #7
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by Moo View Post
    (x-27)-x=x-27-x=(x-x)-27=-27

    And I kept (a-b) in the left side because we don't need to transform them into x.

    I'm so stupid... I read the (x-27)-x as -x(x-27)...

    As I said, I was being dumb
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