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Just need some clarification on radicals

  • 08-01-2013 3:51pm
    #1
    cmssjone
    Registered Users, Registered Users 2 Posts: 363 ✭✭


    Solve for x:
    sqrt(2x + 3) = 2x - 3

    I'm quite happy that I get 2 (possible) solutions of
    x=1/2 and x = 3 and then have to check these in the original radical equation

    Case 1

    x = 3

    Sqrt(9) = 3 implies that x = 3 is a valid solution

    Case 2

    x = 1/2

    sqrt(4) = -2

    To me this implies that x = 1/2 is a valid solution as sqrt(4) is 2 or -2. This is not correct but I am unable to work out why. Any explanation as to why would be much appreciated. Thanks in advance


Comments

  • #2
    hivizman
    Registered Users, Registered Users 2 Posts: 1,163 ✭✭✭hivizman


    If I were solving your equation as an algebraic expression, I'd be inclined to agree with you that there are two solutions, x=3 and x=1/2.

    However, in some contexts, the "square root" function is defined so that only the positive square root is accepted. For example, if you use the "SQRT" function in Excel, this returns only the positive square root. Calculators with square root keys also give only the positive square root.

    So, if your textbook or teacher claims that x=1/2 is not a solution of the equation, check whether they are defining "the square root" as the positive square root and hence ruling out negative numbers from being square roots.


  • #3
    CJC86
    Registered Users, Registered Users 2 Posts: 360 ✭✭CJC86


    hivizman wrote: »
    If I were solving your equation as an algebraic expression, I'd be inclined to agree with you that there are two solutions, x=3 and x=1/2.

    However, in some contexts, the "square root" function is defined so that only the positive square root is accepted. For example, if you use the "SQRT" function in Excel, this returns only the positive square root. Calculators with square root keys also give only the positive square root.

    So, if your textbook or teacher claims that x=1/2 is not a solution of the equation, check whether they are defining "the square root" as the positive square root and hence ruling out negative numbers from being square roots.

    The key is in the word "function". A function can only give one output for a single input, hence we define the square root of a to be the positive root of the equation [latex]x^2 = a[/latex].

    This is not as arbitrary a choice as it appears to be. If we wish to extend the function [latex]f(a) = a^x[/latex] to real numbers rather than just integers, then the only continuous extension is if we select the positive root for each rational number x.

    So, basically there is a big difference between being asked to solve: x = sqrt(a)
    or: [latex]x^2 = a [/latex].


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