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  1. Mar 22, 2015 #1
    Why doesn't x^2+1 has no zeroes i.e, zero of the polynomial??
  2. jcsd
  3. Mar 22, 2015 #2
    ##x^2+1## is always positive. For a negative number -k, you get
    For positive numbers also, it will always return a positive number.
    Since ##x^2>=0##
    Hence it can never be zero.

    {This part is optional:
    If you know calculus,
    For critical points,##2x=0## so x=0 is a point of inflection
    ##d^2y/dx^2=2## which is positive. So at x=0, you have a minimum value.
    So the least possible value attained by the polynomial is 1. }

    Hence, the curve never goes below the y=1.So the polynomial can never be zero.

    Take this as a question for practice:
    Prove that ##x^2-1## has two distinct zeroes
  4. Mar 22, 2015 #3
    Technically, f(x) = x2+1 has no real zeroes. :-p
  5. Mar 22, 2015 #4
    Hey, I have another thought regarding this..
    X=√-1 which is an imaginary number..
    So it can never be zero..
    Last edited: Mar 22, 2015
  6. Mar 22, 2015 #5

    The method with which you tried to show that x^2-1 has two real distinct zeroes is wrong.
    x^2 can also be zero. But it does not have distinct roots.
    For distinct roots, the curve has to come below the x-axis. That is, the parabola should have a negative minimum.
    Or you can use this: b^2 - 4ac>0
  7. Mar 22, 2015 #6
    Sorry.. Whatever the curve, parabola and everything you said, I didn't understand. I haven't took those classes yet. Can you please name that chapter for me??
  8. Mar 22, 2015 #7
    Do you know this formula:
    It is called discriminant.
    When D<0, you have no real roots.
    When D>0, you have 2 distinct roots.
    When D=0, you have 1 root.
  9. Mar 22, 2015 #8
    What is D
  10. Mar 22, 2015 #9
    Do you know how to find roots of a quadratic equation using this formula:
  11. Mar 22, 2015 #10
    Noooo..... I've not taken classes regarding that topic yet.. If you really want to say that to me.. Then probably you must spend all your time here... ;)
  12. Mar 22, 2015 #11
    Last edited: Mar 22, 2015
  13. Mar 22, 2015 #12


    Staff: Mentor

    This is not correct (or at least, not complete).
    If x2 = 1,
    then x2 - 1 = 0
    so (x - 1)(x + 1) = 0
    so x = 1 or x = -1
    The point is, there are two solutions.
    Similar to the above, if x2 + 1 = 0, then x = i or x = -i. Again, there are two solutions, neither of which is real.
    Last edited: Mar 23, 2015
  14. Mar 22, 2015 #13
    Oh yes, I totally missed this point... I got it now.. Thank you Mark44.. :)
  15. Mar 23, 2015 #14
    Hey but its not (x^2-1)^2 ... Its x^2 -1 right?? Then how can it be (x+1)(x-1)
    Last edited by a moderator: Mar 23, 2015
  16. Mar 23, 2015 #15


    Staff: Mentor

    x2 - 1 factors into (x + 1)(x - 1). To verify this for yourself, carry out the multiplication of (x + 1)(x - 1).
  17. Mar 23, 2015 #16
    Yes yes I got it thanks!
  18. Feb 9, 2016 #17
    When D is zero, it doesn't mean there's only one root
    It means there are two roots but both of them are equal!

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