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REALLY NEED HELP Solving for X with radicals/square roots

  1. Oct 13, 2004 #1
    Ok so I know that when you have radicals such as the following you only have one answer right?...


    x + 1 - 2(square root of [x+4]) = 0

    ok so
    x = 5 right?.... i dunno why but it just does... I cant solve it

    If i solve it

    It works looks like this:
    (squaring everything I get this in the end)

    x^2-2x-15=0

    ok so the answers for x = 5 ORRRRRRRRRRRR x = -3????

    why is this so... I must be doing this wrong.. but I dunno how its done=/ plz help.

    if I graph them... the first one just is a l ine that just "ends" =/ and the other one is a parabula of course.
     
  2. jcsd
  3. Oct 14, 2004 #2

    Tide

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    Substitute both values back into the original equation and see what you get!
     
  4. Oct 14, 2004 #3
    I know I can do that... and I know I can prove that one doesnt work BUT is there any other way of doing it.
     
  5. Oct 14, 2004 #4

    Tide

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    The way you solved it is fine but you have to realize that when you squared both sides of the equation you introduced extraneous roots. In the end, the numbers that you find must satisfy the original equation!
     
  6. Oct 14, 2004 #5

    uart

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    If the original equation is to be a function then it can only have one value for each value of x, so that means that the +ive branch of the sqrt is implied.

    The x=-3 is actually a valid solution, but only if you allow the sqrt to return both plus and minus results. You can easily verify that x=-3 is indeed a solution so long as you take the sqrt(1) to be minus 1.

    Be aware however that once you decide to allow for both +ive and -ive branches of that sqrt then the relationship is multi-valued and at each of the solution points (x-5 and x=-3) you have both x + 1 - 2 sqrt(x+4) = 0 and simultaneously x + 1 - 2 sqrt(x+4) != 0. This is because the relation returns two results for each value of x, one result that is zero and one result that is not.
     
    Last edited: Oct 14, 2004
  7. Oct 14, 2004 #6
    Ok.. I cant really think about it over right now..

    but ok what about absolute values

    example abs value of (3x-2) = 2 TIMES sqr of (x+8)

    How do i do that?
     
  8. Oct 14, 2004 #7
    How about when you have just abs values on two sides of an equation?

    What about when you have sqr roots of x's on both sides?
     
  9. Oct 14, 2004 #8

    Tide

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    Solve separately for the two situations where 3x-2 > 0 and 3x-2 < 0 and be sure your solutions comform to those requirements.
     
    Last edited: Oct 14, 2004
  10. Oct 14, 2004 #9
    abs(x) = sqrt(x^2) therefore abs(3x-2) = sqrt((3x-2)^2), this gives

    sqrt((3x-2)^2) = 2*sqrt(x+8)
    (3x-2)^2 = 4*(x+8)
    This equation you can probably solve, be sure to check the roots ...
     
  11. Oct 14, 2004 #10

    HallsofIvy

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    In general, if you square both sides of an equation or multiply both sides of an equation by something involving the unknown x, you may introduce "extraneous roots": values that satisfy the new equation but not the original one.


    Obvious example: the equation x= 3 has only 3 as a root but if we multiply both sides by x-2, x(x-2)= 3(x-2) has both 3 and 2 as roots.
     
  12. Oct 14, 2004 #11
    well i still dont fully understand it all. so does anyone have a guide that explains why only one of the answers is correct on the original problem?


    Something that talks about having abs value on both sides of the equation woudl also be helpful thanks.
     
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