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Solving quadratics and factorisation of polynomials using calculus

  1. Feb 3, 2005 #1
    I just found this really old book. In it, I found a way of solving quadratic equations using calculus. Ive never seen this method in any other book. Ok, heres the method :

    The discriminant of the quadratic formula i.e sqrt(b^2 - 4ac) is equal to the first derivative of the original quadratic expression. I am not sure how this is derived because the author doesnt give a derivation.

    Theres also one simple relation between the first derivative of an expression and the factors of it in the form (x-a)(x-b)(x-c).... which is : The sum of the factors is equal to the first derivative. This is easy to prove but i have never seen this in any other calculus book or algebra book. Heres the proof :

    Lets use a quadratic expression for simplicity,y, and its factors are (x+a)(x+b)

    (d/dx)y = (x+a)((d/dx)*(x+b)) + (x+b)((d/dx)*(x+a)
    = x+a + x+b

    Using this, you can find the factors from the derivative or viceversa.

    I think the first method I described can also be derived in a similar way, but the author didnt do it in the book.
  2. jcsd
  3. Feb 4, 2005 #2


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    I think you are leaving out a crucial piece of information. Obviously, the derivative of a quadratic function is not a constant.
  4. Feb 4, 2005 #3


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    Was it printed before 1650?? :confused: :tongue2:

  5. Feb 4, 2005 #4
    No, it was written in 1940, and printed in 1960
  6. Feb 4, 2005 #5


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    lol...are you trying to say (in a tongue in cheek way) that nobody knew about "the calculus" before 1650, and since it seems to you that the author of this book was writing a load of bull****, that he certainly didn't know anything about calculus, so it's as though his work dates back to a more naive time?

    Or did I misinterpret the joke entirely?
  7. Feb 4, 2005 #6


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    Daniel is deep, cepheid. You cannot fathom him.

    Apart from that, WORLD-HEN:
    It makes absolutely no sense to say that the first derivative of a second-degree polynomial equals a constant.
    So, you have misunderstood something!

    The "sum of factors" you're talking about, is called the "trace".
  8. Feb 4, 2005 #7


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    Slow down fellow's!

    We have:

    [tex] f(x) = ax^2+ bx + c =0 [/tex]

    is satisfied by

    [tex] x = \frac { -b \pm \sqrt { b^2 - 4ac}} {2a} [/tex]

    so it follows that

    [tex] 2ax = -b \pm \sqrt { b^2 - 4ac} [/tex]

    [tex] f'(x) = 2ax + b = \pm \sqrt { b^2 - 4ac} [/tex]

    of course this would only hold for [itex] x [/itex] s.t. [itex] f(x) =0 [/itex]

    Looks to me like his claim is a result of simple algebra. It is interesting that the slope of a quadratic at a root is given by the determinate.

    Now what use is this?
  9. Feb 4, 2005 #8


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    Been thinking about this. Back before graphing calculators, one of the things you often did was to create a quick sketch of a function without having to plot very many points. You could find the roots, the max and mins and then just free hand the function through those points. This shows that if you have used the quadratic formula to compute the roots, you have also got the SLOPE of the function at the roots. If you have ever done free hand function sketches I think you will agree that this is a nice bit of information. To bad it is only good for quadratics.
    Last edited: Feb 4, 2005
  10. Feb 4, 2005 #9
    Here's the algebraic form of calculus (though a lot less impressive, and more tedious)

    we are trying to find the derivative at (1,12) on the graph x^2+6x+5=y

    where x has only one solution, we are trying to solve for m.

    m(1)+b=12 b=12-m
    we need the discriminant equal to zero to have one solution
    (m-8)(m-8)=0 here m=8
  11. Feb 4, 2005 #10
    i suppose that on a cubic graph the tangent line would create two solutions.
  12. Feb 4, 2005 #11
    Well, the matter is really very simple, no derivative needed. Let[tex] \alpha, \beta [/tex]be the roots.

    By the nature of the polynomial [tex]X^2+(B/A)X + C/A = (X-\alpha)(X-\beta)=X^2-(\alpha +\beta)X+\alpha\beta [/tex].

    Thus -B/A =[tex] (\alpha+\beta); C/A =\alpha\beta.[/tex]

    [tex](-B/A)^2-4C/A =(\alpha-\beta)^2[/tex]

    Thus the quadratic formula separates out the roots,using first the plus sign:

    [tex]-B/A + \frac{\sqrt{b^2-4AC}}{A} = 2\alpha.[/tex]

    So the sum of the roots is found from the coefficients.
    Last edited: Feb 4, 2005
  13. Feb 6, 2005 #12


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    Very cool, and I agree useful for drawing parabolas by hand. It makes perfect sense: if the discriminant is zero, that slope is zero...which is exactly what happens when you have just one root...the graph intersects the x-axis at the vertex only. If the discriminant is positive, you have two slopes, one +ve on -ve, which correspond to the slopes of the graph at each of the two roots. If the discriminant is negative, the roots are complex. So with no real roots, I guess it's meaningless to talk of the slope of the graph there!
  14. Feb 6, 2005 #13
    Actually , you can extend it to all polynomials.

    for a cubic equation with factors a,b,c

    The second derivative = 2!(a + b + c)

    for a biquadratic expression with factors a,b,c,d
    The thirds derivative = 3!(a + b + c + d)

    Similarly, for a n-nomial
    the (n-1)th derivative = (n-1)!( a + b + c + ....)
  15. Feb 6, 2005 #14


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    thanks for the clever ideas. I thought the original post was nutty until integral slowed us down. As several of you have already pointed out, what is going on is a combination of two ancient algebraic ideas: first, the derivative is the slope of the line which intersects the curve doubly at the point, so that makes the difference of the two equations have a double root, i.e. makes the discriminant zero. this is merely descartes method of finding slopes of tangent lines, the algebraic form of calculus which greatly precedes newtons limiting method.

    this old idea survives today as the zariski tangent space definition in abstract algebraic geometry.

    from this point of view to find the tangent slope of f(x) a polynomial, one seeks the number m such that f(x)-(f(a) + m(x-a)) has a double root at x=a. i.e. such that the discriminant is zero. I.e. since this expression already has one root at x=a, one wants to divide out this root and them have another.

    Dividing out gives the expression [f(x)-f(a)]/(x-a) - m. I.e. if f(x) is a polynomial then x-a divides into f(x)-f(a) evenly say, giving f(x)-f(a) = g(x)(x-a).

    so to have [f(x)-f(a)]/(x-a) - m = g(x) - m equal zero at x=a, one needs of course to have m = g(a), where g(x) = [f(x)-f(a)]/(x-a). This is exactly newton's definition of the derivative of f at a. i.e. to evaluate the limit of [f(x)-f(a)]/(x-a) at x=a, one can first remove the discontinuity at x=a, by replacing the quotinet [f(x)-f(a)]/(x-a) by g(x), and then simply evaluate at x=a.

    now what does this have to do with discriminants? well the original definition of a (squared) discriminant is as the product of the squares of the pairwise differences of all roots. this is why it equals zero if two roiots are the same.

    then this is clearly a symmetric function oif the roots,m hence can be expressed as a polynomikla in the elementary symmetric functions, which are merely the coefficients opf the polynomial, e.g. b^a - 4ac for a quadratic.

    as integrtal has pointed out, the squared discriminant easily gives the square of the slope at a root. now any f of course is normally transformed into one with a root at the interesting point by forming instead f(x) - f(u). so these two polynomials have thje same slope, and one of them has a root at the interesting point u. so one needs to relate the discriminants of f(x) and f(x)-f(u). but if we divide out f(x)-f(u) by x-u, we get for the other root (lets asume the leading coefficient a = 1, since this really is the egneral case) then the other root is -u-b. forming the discriminant by the classical definition we subtract u from this elaving -2u-b, and squaring gives the square of the original derivative. (2u+b)^2.

    so indeed the discriminant of the modified quadratic polynomial f(x)-f(u), which integral shoiwed equals the slope of the modified polynomial, also equals the slope of the original one.

    Interestingly too, if we form the further modified polynomial f(x)-f(u) - m(x-u), the polynomila which has a double root at u if and only if m = slope of f at u, we see that the discriminant of f(x)-f(u) - m(x-u) equals zero if and only if m = slope of f at u.

    Indeed calculation shows that the (squared) discriminant of f(x)-f(u) - m(x-u), is just (m - f'(u))^2. where f'(u)^2 is the (squared) discriminant of f(x)-f(u).

    Last fall I taught descartes method for derivatives to my class, as more intuitive than newtons and easier to understand. indeed newton's methois is not needed until one encounters the first case where the division of f(x)-f(u) by (x-u) cannot be carried out, i.e., for sin(x).

    Kudoes to everyone here who persisted when a post seemed to be non sense, especially Integral. (Oh yes, for some of these calculations one needste diea given aboive that the sum of the roots is minus the linear coefficient of a monic quadratic, i.e. the coefficients are really the elementary symmetric functions of the roots.)
  16. Feb 6, 2005 #15


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    by the way, descartes method for finding slopes occurs in the very obscure book, lectures on freshman calculus, by cruse and granberg, which i have recommended here before, although the authors state incorrectly that the method works only for quadratics. (I tried to correct the error as a paid "reviewer" but was ignored.)
    Last edited: Feb 6, 2005
  17. Feb 6, 2005 #16


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    so to summarize, the discriminant tells when a polynomial has a repeated root, and descartes method for slopes says that m is the slope of y = f(x) at x=a, if and only if the polynomial f(x)-f(a) - m(x-a) has a repeated root at x=a, if and only if g(a) = m, where f(x) = (x-a)g(x).

    this combined with the algebraic facts that the (squared) discriminant is the square of the product of the pairwise differences of the roots, and the fact that the second coefficient of a monic polynomial is (minus) the sum of the roots, explains all the phenomena above.

    these results are ancient, and are among those which led fermat and descartes to the computation of slopes for polynomials before newton and leibniz. indeed the limiting method of newton is merely a technical trick for evaluating [f(x)-f(a)]/(x-a) at the point x=a, when we cannot directly divide it out.

    moreover the method of descartes survives as the definition in algebraic geometry of the zariski cotangent space of an algebraic variety at a point p, i.e. as M/M^2, where M is the maximal ideal of functions regular at p and vanishing at p, and M^2 is therefore the ideal of functions vanishing twice at p. Hence M/M^2 is the vector space of functions vanishing at p, modulo those that vanish twice, i.e. modulo those which not only vanish at p but also have derivative zero at p. Hence the quotient space M/M^2 is the vector space of all (directional) derivatives at p, i.e. the dual space of the space of all directions at p.

    from a more concrete point of view, imagine a function f which vanishes at p expanded in a taylor series, and then throw away all the terms of order two or higher. what is left is the derivative of f, i.e. the linear term of the taylor series, i.e. the part of the function which is left when we mod out by the part vanishing "twice" (or more).

    Hence M/M^2 is called the Zariski cotangent space and its dual (M/M^2)* is called the Zariski tangent space.

    Of course there are no dot products intrinsically present here so one must clearly distinguish between tangent spaces and their duals. :biggrin:

    by the way, to take the derivative of f this way, one merely passes from f to the equivalence class of f-f(p) in the quotient space M/M^2. i.e. [f-f(p)] = dfp, in M/M^2.

    sheaf theory, meet fermat. i.e. "there is nothing new under the sun."
    Last edited: Feb 6, 2005
  18. Feb 6, 2005 #17
    So you find the vertex, and you know the shape of the parabola based on the coefficient of the x^2 term. And do the sqrt of the y coordinate of the vertex from below the x-axis. And then do the x coordinate of the vertex plus and minus the sqrt of the y coordinate of the vertex. that should give the root, of a quadratic with the coeffient of the x^2 term being one. Can we use the same method on the cubic?, the problem is it that it doesnt have symmetry like a quadratic
  19. Feb 7, 2005 #18


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    the descartes method for finding the derivative of a general polynomial f is to solve for the unique number m such that the polynomial [f(x)-f(a) - m(x-a)] has a double root at x=a.

    by the root factor theorem, since x=a is a root of f(x)-f(a), one can factor

    f(x)-f(a) as (x-a)g(x) for some polynomial g. then we have [f(x)-f(a) - m(x-a)]

    = (x-a)[g(x) - m], hence f(x)-f(a) has a double root at x=a if and only if x=a is a root of g(x) - m. i.e. if and only if m = g(a).

    thiods is exactly newtons answer/

    e.g. trying it on f(x) = x^n, gives g(a) = n a^(n-1).

    i.e. x^n - a^n = (x-a)[x^(n-1) + ax^(n-2) + ...+a^(n-1)] so g(a) = n a^(n-1).

    it is not so easy to use only the discriminant in higher dergees, since it does not detect where the double root occurs, which is trivial for quadratics.
  20. Feb 7, 2005 #19
    that certainly is the problem, because there is no easy way of detecting the discriminant of higher degrees. Though i dont think it is absolutly necessary for one to detect to the discriminant for higher degrees. I think that its easier to see where a cubic and a line have two roots rather than one or three. and then knind of factor your way to the answer.
    For example, given a cubic polynomial:
    trying to find the derivative at (1, -12)
    where x has only two distinct solutions, then it must have a unique m.
    so this can factor to (x-a)(x-a)(x-b)
  21. Feb 7, 2005 #20
    a little mess
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