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Thread: Solve Y= (4th root of x) +1

  1. #1

    Solve Y= (4th root of x) +1

    Solve y =(4th root of x) +1

    I have graphed this function and can see that there are no real solutions for x. I would draw the graph here but don't know how to. It starts off at (0,1) and the x value increases rapidly as the y value increases slowly.
    But then I tried to solve for x using algebra and found a real solution.
    See below:

    Let 0=(4th root of x) +1
    -1=(4th root of x)
    (-1)^4 = (4th root of x)^4
    1 = x

    So I have a solution of x =1 and yet the function does not cross the x axis at any point when I graph it.

    Is someone able to show me why there is no solution on the graph but there is one when I use algebra. I am assuming that my algebra is wrong somewhere but can't see where.

    Thanks a lot.

    Zen

  2. #2
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    Quote Originally Posted by Archie Deetoo View Post
    Solve y =(4th root of x) +1

    I have graphed this function and can see that there are no real solutions for x. I would draw the graph here but don't know how to. It starts off at (0,1) and the x value increases rapidly as the y value increases slowly.
    But then I tried to solve for x using algebra and found a real solution.
    See below:

    Let 0=(4th root of x) +1
    -1=(4th root of x)
    (-1)^4 = (4th root of x)^4
    1 = x

    So I have a solution of x =1 and yet the function does not cross the x axis at any point when I graph it.

    Is someone able to show me why there is no solution on the graph but there is one when I use algebra. I am assuming that my algebra is wrong somewhere but can't see where.
    Let's be clear about terminology. When you talk about "solving", I think you mean finding zeros (or y-intercepts) of the function y = x^(1/4) + 1, that is, solutions of the equation x^(1/4) + 1 = 0. You are right that this equation has no real solutions.

    What you did in solving is to raise both sides to the 4th power, which introduces extraneous solutions. You have to check, and you find that x = 1 is not in fact a solution of the original equation.

  3. #3
    Senior Member
    Join Date
    Dec 2014
    Posts
    2,429
    Quote Originally Posted by Archie Deetoo View Post
    Solve y =(4th root of x) +1

    I have graphed this function and can see that there are no real solutions for x. I would draw the graph here but don't know how to. It starts off at (0,1) and the x value increases rapidly as the y value increases slowly.
    But then I tried to solve for x using algebra and found a real solution.
    See below:

    Let 0=(4th root of x) +1
    -1=(4th root of x)
    (-1)^4 = (4th root of x)^4
    1 = x

    So I have a solution of x =1 and yet the function does not cross the x axis at any point when I graph it.

    Is someone able to show me why there is no solution on the graph but there is one when I use algebra. I am assuming that my algebra is wrong somewhere but can't see where.

    Thanks a lot.

    Zen
    Zen, when you raise numbers to an even power you might get erroneous roots.
    Consider the equation x=-2 (SO X EQUALS -2 AN NOTHING ELSE)
    Now square both sides and get x^2 =4
    Solving x^2=4 we get x = +/- 2. Clearly x=2 is an erroneous solution!
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  4. #4
    Quote Originally Posted by Dr.Peterson View Post
    Let's be clear about terminology. When you talk about "solving", I think you mean finding zeros (or y-intercepts) of the function y = x^(1/4) + 1, that is, solutions of the equation x^(1/4) + 1 = 0. You are right that this equation has no real solutions.

    What you did in solving is to raise both sides to the 4th power, which introduces extraneous solutions. You have to check, and you find that x = 1 is not in fact a solution of the original equation.
    Thank you very much, Doctor. Great help.

  5. #5
    Quote Originally Posted by Jomo View Post
    Zen, when you raise numbers to an even power you might get erroneous roots.
    Consider the equation x=-2 (SO X EQUALS -2 AN NOTHING ELSE)
    Now square both sides and get x^2 =4
    Solving x^2=4 we get x = +/- 2. Clearly x=2 is an erroneous solution!
    Thanks a lot for your help. Makes sense. Cheers!

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