My Math Forum Question about proving the availability of double real root of a polynomial

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 October 31st, 2018, 02:50 AM #1 Member   Joined: Nov 2012 Posts: 74 Thanks: 1 Question about proving the availability of double real root of a polynomial 23A5BCBB-CA79-4DA7-B726-23E5A84070E1.jpg 2BC82CF6-7D69-4698-87BF-11388DD9C0A7.jpg E2282EA8-2E2A-45F4-B13F-D9D0ACAFB0B4.jpg For part b, it is assuming that n is an odd number, so r-a = r+a Since the exponent of n-1 would be an even number for n being an odd number, is the equation of (r-a)^(n-1) = -(r+a)^(n-1) undefined? It is because any real number to a power of an even number cannot be negative. Please rectify me if I have made some mistakes.
 October 31st, 2018, 09:07 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,250 Thanks: 516 I can't read your attachments so I cannot actually check your work. $f(x) \text { is a polynomial such that } f(x) = (x - r)^2 * g(x) \text { and } g(r) \ne 0.$ $f'(x) = (x - r)^2 * g'(x) + 2(x - r) * g(x) \implies f'(r) = 0 * g'(r) + 2 * 0 * g(r) = 0.$ $p(x) = (x - a)^n + (x + a)^n = \displaystyle = \left ( \sum_{k=0}^n \dbinom{n}{k} * (-\ 1)^k * x^{(n-k)}a^k \right ) + \left ( \sum_{k=0}^n \dbinom{n}{k} * x^{(n-k)}a^k \right ).$ $n = 2m \implies p(x) = \displaystyle \left ( \sum_{j=0}^m 2 * \dbinom{2m}{2j} * (x^{(m - j)})^2 * (a^j)^2 \right ) > 0 \ \because a \ne 0 \text { by hypothesis.}$ Every term in that sum is non-negative except the last, which is positive, so the sum is positive. There are no real roots, let alone a real double root. We skip the trivial case where n = 1. $p'(x) = \displaystyle \left ( \sum_{k=0}^{n-1} (n - k) \dbinom{n}{k} * (-\ 1)^k * x^{(n - 1 - k)}a^k \right ) + \left ( \sum_{k=0}^{n - 1} (n - k) * \dbinom{n}{k} * x^{(n-1-k)}a^k \right ).$ $n = 2m + 1 \implies p'(x) = \displaystyle \left ( \sum_{j=0}^m 2 * (2m + 1 - 2j) \dbinom{2m + 1}{2j} * (x^{(m - j)})^2(a^j)^2 \right ).$ $\therefore n = 2m + 1 \text { and } x \ne 0 \implies p'(x) > 0.$ $n = 2m + 1 \text { and } p'(x) = 0 \implies x = 0.$ $p(0) = a \ne 0.$ If n is odd, the derivative is not zero at any real root, and therefore no real root is a double root.
 October 31st, 2018, 08:40 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,098 Thanks: 1905 Using latex: Let $f(x)$ be a polynomial. If $f(x) = (x - r)^2g(x)$ where $g(r) \ne 0$, we say that $r$ is a double root of $f(x) = 0$. (a) $\$ If $f(x) = 0$ has a double root, prove that $f(r) = f'(r) = 0$. (b) $\$ Show that $p(x) = (x - a)^n + (x + a)^n$, where $a$ is a non-zero real number, does not have a double real root. 2. $\ \$ (a) $\$ If $f(x) = 0$ has a double root $r$, \displaystyle \begin{align*} f(x) &= (x - r)^2g(x) \\ \hspace{80px}f(r) &= (r - r)^2g(r) \\ &= 0 \\ f'(x) &= \frac{d}{dx}[(x - r)^2g(x)] \\ &= (x - r)^2\frac{d}{dx}[g(x)] + g(x)\frac{d}{dx}(x - r)^2 \\ &= (x - r)^2g'(x) + g(x)(2)(x - r)\frac{d}{dx}(x - r) \\ &= (x - r)^2g'(x) + 2(x - r)g(x) \\ f'(r) &= (r - r)^2g'(r) + 2(r - r)g(r) \\ &= 0\end{align*} $\hspace{80px}\therefore$ If $f(x) = 0$ has a double root $r$, then $\hspace{100px}f(r) = f'(r) = 0$ $\hspace{28px}$(b) $\$ Assume $p(x)$ has a double real root $r$, \displaystyle \hspace{66px}\therefore\ p(r) = p'(r) = 0 \\ \begin{align*}\hspace{66px}p'(x) &= \frac{d}{dx}[(x - a)^n + (x + a)^n] \\ &= n(x - a)^{n-1} + n(x + a)^{n-1} \\ \end{align*} \displaystyle \begin{align*}\hspace{100px}p'(r) &= n(r - a)^{n-1} + n(r + a)^{n-1} = 0 \\ (r - a)^{n-1} &= -(r + a)^{n-1}\end{align*} Thanks from justusphung
 November 1st, 2018, 02:49 AM #4 Member   Joined: Nov 2012 Posts: 74 Thanks: 1 I am quite confused with the binomial expression, though I think, for the p(x), you are proving that the graph does not cut the x-axis given that a is not equal to zero. Therefore, the graph has no any real roots at all. Anyway, thx for your help Jeff. I checked up the answer of the question again. The derivative of p(x) is n(x+a)^(n-1) + n(x-a)^(n-1) = 0 by assuming that there is a double real root. We can re-write the equation to the form of (x-a)^(n-1) = -(x+a)^(n-1). Before, I was wondering why if n-1 was an even number, then x-a = x+a. Someone just told me that only when x-a = x+a = 0, the above equation would be valid. And now, I am convinced that x-a = x+a when n-1 is an even number. This leads to the result that a=0 which contradicts the fact of the constant a being an non-zero real number to disprove the assumption for n-1 being an even integer.
November 1st, 2018, 06:38 AM   #5
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 Originally Posted by justusphung I am quite confused with the binomial expression, though I think, for the p(x), you are proving that the graph does not cut the x-axis given that a is not equal to zero. Therefore, the graph has no any real roots at all. Anyway, thx for your help Jeff. I checked up the answer of the question again. The derivative of p(x) is n(x+a)^(n-1) + n(x-a)^(n-1) = 0 by assuming that there is a double real root. We can re-write the equation to the form of (x-a)^(n-1) = -(x+a)^(n-1). Before, I was wondering why if n-1 was an even number, then x-a = x+a. Someone just told me that only when x-a = x+a = 0, the above equation would be valid. And now, I am convinced that x-a = x+a when n-1 is an even number. This leads to the result that a=0 which contradicts the fact of the constant a being an non-zero real number to disprove the assumption for n-1 being an even integer.
First, skipjack has provided a far more elegant proof than mine, and I think you have grasped it.

Second, I think you misunderstood mine. My proof is split into 2 parts.

If n is even, p(x) has no real root and therefore no real double root. For example

$p(x) = (x - a)^2 + (x + a)^2 = x^2 - 2ax + a^2 + x^2 - 2ax + a^2 = 2x^2 + 2a^2.$

And $2x^2 + 2a^2 > 0 \text { unless } x = 0 = a.$

$\text {But } a \ne 0 \text { by hypothesis.}$

So no real root and thus no double real root.

It is a bit more complex if n is odd. Then p(x) certainly has at least one real root. But p'(x) has no real root. But that is a requirement for a double real root. For example

$p(x) = (x - a)^3 + (x + a)^3 =$

$x^3 - 3ax^2 + 3a^2x - a^3 + x^3 + 3ax^2 + 3a^2x + a^3 = 2x^3 + 6a^2x \implies$

$p'(x) = 6x^2 + 6a^2 > 0 \text { unless } x = 0 = a.$

$\text {But } a \ne 0 \text { by hypothesis.}$

So no value of x can be a double root because p'(x) would have to equal 0 there.

Last edited by JeffM1; November 1st, 2018 at 06:41 AM.

 November 2nd, 2018, 11:41 PM #6 Member   Joined: Nov 2012 Posts: 74 Thanks: 1 Really appreciate for your work, Jeff. It seems that it is more convenient for me to understand the steps by showing some counter-examples to disprove the assumption rather than by showing in the form of algebraic expressions. Really thanks for your time And also jack for your suggestion. Last edited by justusphung; November 3rd, 2018 at 12:04 AM.
 November 3rd, 2018, 08:14 AM #7 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,628 Thanks: 119 p(x)=(x-a)^n+(x+a)^n=0 If n even, impossible. If n odd (a-x)^n=(a+x)^n $\displaystyle \rightarrow$ a-x=a+x $\displaystyle \rightarrow$ x=-x x=0 is only real root

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