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Math Help - Help me check this quadratic equation

  1. #1
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    Help me check this quadratic equation

    \frac{x^2+1}{x}+\frac{x}{x^2+1}=2.9
    Substitution: z=\frac{x^2+1}{x}
    z+\frac{1}{z}=2.9
    z^2-2.9z+1=0
    z_1,_2=\frac{2.9\mp\sqrt{4.41}}{2}
    z_1=2.5
    z_2=0.4

    So

    2.5=\frac{x^2+1}{x}
    x_1,_2=\frac{2.5\mp\sqrt{2.25}}{2}
    x_1=2
    x_2=0.5

    and

    0.4=\frac{x^2+1}{x}
    x^2-0.4x+1=0
    x_3,_4=\frac{0.4\mp\sqrt{0.16-4}}{2}

    Which brings us to no solution

    Is this right?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nathan02079 View Post
    \frac{x^2+1}{x}+\frac{x}{x^2+1}=2.9
    Substitution: z=\frac{x^2+1}{x}
    z+\frac{1}{z}=2.9
    z^2-2.9z+1=0
    z_1,_2=\frac{2.9\mp\sqrt{4.41}}{2}
    z_1=2.5
    z_2=0.4

    So

    2.5=\frac{x^2+1}{x}
    x_1,_2=\frac{2.5\mp\sqrt{2.25}}{2}
    x_1=2
    x_2=0.5

    and

    0.4=\frac{x^2+1}{x}
    x^2-0.4x+1=0
    x_3,_4=\frac{0.4\mp\sqrt{0.16-4}}{2}

    Which brings us to no solution

    Is this right?
    i did not work it out, but i believe it has a solution. your z_2 should be {\color {red} -} 0.4

    EDIT: in fact, the solutions are x = 1/2 and x = 2
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    i did not work it out, but i believe it has a solution. your z_2 should be {\color {red} -} 0.4

    EDIT: in fact, the solutions are x = 1/2 and x = 2
    but isnt
    z_2=\frac{2.9-2.1}{2}?
    then
    z_2=\frac{0.8}{2}
    which is then
    z_2=0.4

    well at least thats how i came up with the answer:P
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nathan02079 View Post
    but isnt
    z_2=\frac{2.9-2.1}{2}?
    then
    z_2=\frac{0.8}{2}
    which is then
    z_2=0.4

    well at least thats how i came up with the answer:P
    yes. you are correct, my bad. i did 2.1 - 2.9 instead of 2.9 - 2.1

    so you made a mistake somewhere else then... let's see...ah, no mistake. you got the two solutions i gave! x_1,2 is your answer. the others are complex simply because no (real) solution exists for that value of z. it's fine
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