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Thread: Show that the equation , has no real roots for all real values of ݇k

  1. #1
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    Show that the equation , has no real roots for all real values of ݇k

    Hi I am new here I hope I am posting in the right section

    I am having some trouble with this sum I was hoping to get some help here.

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  2. #2
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    Re: Show that the equation , has no real roots for all real values of ݇k



    check the discriminant ...

    $k^2 - 4(1)(k^2+2) = -(3k^2+8) < 0 \implies y = x^2+kx+(k^2+2)$ has no real zeros.
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    Re: Show that the equation , has no real roots for all real values of ݇k

    may I ask how you got the numbers -4,1,3 and 8?
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    Re: Show that the equation , has no real roots for all real values of ݇k

    Quote Originally Posted by Taine145 View Post
    may I ask how you got the numbers -4,1,3 and 8?
    The discriminant is the part that goes inside the square root from the Quadratic Formula:

    $x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

    The discriminant is just $b^2-4ac$, which in the case of your quadtratic, you have $a=1, b=k, c=k^2+2$. Plug in, and see that for any real number $k$, you wind up with a negative discriminant (so when you take the square root, you get an complex number for the root).
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    Re: Show that the equation , has no real roots for all real values of ݇k

    Another way- completing the square:
    x^2+ kx+ k^2+ 2= x^2+ kx+ \frac{k^2}{4}- \frac{k^2}{4}+ k^2+ 2= (x+ \frac{k}{2})^2- \frac{k^2}{4}+ \frac{4k^2}{4}+ \frac{8}{4}= (x+ \frac{k}{2})^2+ \frac{3k^2+ 8}{4}= 0

    That is the same as (x+ \frac{k}{2})^2= -\frac{3k^2+ 8}{4}. Since a square (of a real number) is never negative, that is possible only as long as [tex\-\frac{3k^2+ 8}{4}[/tex] is positive. But that in turn is true only if 3k^2+ 8 is negative which is not true for any k because 3k^2 is non-negative and 8 is positive.
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    Re: Show that the equation , has no real roots for all real values of ݇k

    I am more familiar with this way, Thank you guys I really apprciate it
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    Re: Show that the equation , has no real roots for all real values of ݇k

    As stated previously, $ax^2+bx+c = 0 \implies x = \dfrac{-b \pm \sqrt{\color{red}{b^2-4ac}}}{2a}$

    The discriminant, $\color{red}{b^2-4ac}$, is a powerful tool to analyze quadratics ...

    If $b^2-4ac < 0$, the quadratic has no real zeros

    If $b^2-4ac = 0$, the quadratic has one real zero of multiplicity two

    If $b^2-4ac > 0$, the quadratic has two distinct, real zeros ...

    furthermore, if the discriminant is a perfect square value, the quadratic will factor over the rational numbers ... nice to know if you want to factor the quadratic.
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