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October 31st, 2018, 02:50 AM  #1 
Member Joined: Nov 2012 Posts: 65 Thanks: 1  Question about proving the availability of double real root of a polynomial 23A5BCBBCA794DA7B72623E5A84070E1.jpg 2BC82CF67D69469887BF11388DD9C0A7.jpg E2282EA82E2A45F4B13FD9D0ACAFB0B4.jpg For part b, it is assuming that n is an odd number, so ra = r+a Since the exponent of n1 would be an even number for n being an odd number, is the equation of (ra)^(n1) = (r+a)^(n1) 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: 1,192 Thanks: 489 
I can't read your attachments so I cannot actually check your work. $f(x) \text { is a polynomial such that } f(x) = (x  r)^2 * g(x) \text { and } g(r) \ne 0.$ $f'(x) = (x  r)^2 * g'(x) + 2(x  r) * g(x) \implies f'(r) = 0 * g'(r) + 2 * 0 * g(r) = 0.$ $p(x) = (x  a)^n + (x + a)^n = \displaystyle = \left ( \sum_{k=0}^n \dbinom{n}{k} * (\ 1)^k * x^{(nk)}a^k \right ) + \left ( \sum_{k=0}^n \dbinom{n}{k} * x^{(nk)}a^k \right ).$ $n = 2m \implies p(x) = \displaystyle \left ( \sum_{j=0}^m 2 * \dbinom{2m}{2j} * (x^{(m  j)})^2 * (a^j)^2 \right ) > 0 \ \because a \ne 0 \text { by hypothesis.}$ Every term in that sum is nonnegative 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. $p'(x) = \displaystyle \left ( \sum_{k=0}^{n1} (n  k) \dbinom{n}{k} * (\ 1)^k * x^{(n  1  k)}a^k \right ) + \left ( \sum_{k=0}^{n  1} (n  k) * \dbinom{n}{k} * x^{(n1k)}a^k \right ).$ $n = 2m + 1 \implies p'(x) = \displaystyle \left ( \sum_{j=0}^m 2 * (2m + 1  2j) \dbinom{2m + 1}{2j} * (x^{(m  j)})^2(a^j)^2 \right ).$ $\therefore n = 2m + 1 \text { and } x \ne 0 \implies p'(x) > 0.$ $n = 2m + 1 \text { and } p'(x) = 0 \implies x = 0.$ $p(0) = a \ne 0.$ 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: 19,878 Thanks: 1834 
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 nonzero 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 $r$, $\displaystyle \hspace{66px}\therefore\ p(r) = p'(r) = 0 \\ \begin{align*}\hspace{66px}p'(x) &= \frac{d}{dx}[(x  a)^n + (x + a)^n] \\ &= n(x  a)^{n1} + n(x + a)^{n1} \\ \end{align*}$ $\displaystyle \begin{align*}\hspace{100px}p'(r) &= n(r  a)^{n1} + n(r + a)^{n1} = 0 \\ (r  a)^{n1} &= (r + a)^{n1}\end{align*}$ 
November 1st, 2018, 02:49 AM  #4 
Member Joined: Nov 2012 Posts: 65 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 xaxis 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)^(n1) + n(xa)^(n1) = 0 by assuming that there is a double real root. We can rewrite the equation to the form of (xa)^(n1) = (x+a)^(n1). Before, I was wondering why if n1 was an even number, then xa = x+a. Someone just told me that only when xa = x+a = 0, the above equation would be valid. And now, I am convinced that xa = x+a when n1 is an even number. This leads to the result that a=0 which contradicts the fact of the constant a being an nonzero real number to disprove the assumption for n1 being an even integer. 
November 1st, 2018, 06:38 AM  #5  
Senior Member Joined: May 2016 From: USA Posts: 1,192 Thanks: 489  Quote:
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: 65 Thanks: 1 
Really appreciate for your work, Jeff. It seems that it is more convenient for me to understand the steps by showing some counterexamples 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: 1,603 Thanks: 115 
p(x)=(xa)^n+(x+a)^n=0 If n even, impossible. If n odd (ax)^n=(a+x)^n $\displaystyle \rightarrow$ ax=a+x $\displaystyle \rightarrow$ x=x x=0 is only real root 

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availability, double, polynomial, proving, question, real, root 
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