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Let $r_1,r_2,r_3,\cdots,r_n$ be the distinct real zeroes of the equation $$x^8-14x^4-8x^3-x^2+1=0.$$ Then $r_1^2+r_2^2+r_3^2+\cdots+r_n^2$ is $$(A)\,3\quad(B)\,14\quad(C)\,8\quad(D)\,16$$

I can get the sum of the squares of all roots using Vieta’s formulae, but I don't know actually how to proceed in this question.

Do I need to draw a graph and then find the answers or is there a sum trick here which I am not able to see through?

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  • $\begingroup$ Once you get Vieta's formulae you can continue with Newton's identity, basically $a_ns_2+a_{n-1}s_1+a_{n-2}=0$. But as dxiv said, you have first to isolate the polynomial with real roots. $\endgroup$ – zwim Sep 17 '18 at 19:44
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Hint:   the real roots are the roots of the first factor:

$$ \begin{align} x^8\color{red}{+2x^4-2x^4}-14x^4-8x^3-x^2+1 &= (x^4+1)^2 - x^2(4x+1)^2 \\ &= (x^4 - 4 x^2 - x+1)(x^4+4x^2+x+1) \end{align} $$

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  • $\begingroup$ but how is it guaranteed that all roots of first factor are real? $\endgroup$ – maveric Sep 17 '18 at 20:42
  • $\begingroup$ @ashishdeosingh By inspection, it is positive at $\,x=0\,$ and negative at $\,x=\pm 1\,$. $\endgroup$ – dxiv Sep 17 '18 at 20:45
  • $\begingroup$ that is we using IVT. right. will it be possible for us to factorise it? $\endgroup$ – maveric Sep 17 '18 at 20:49
  • $\begingroup$ @ashishdeosingh Using IVT you can tell that there is one real root in each of $\,(-\infty, -1)\,$, $\,(-1,0)\,$, $\,(0,1)\,$, $\,(1,\infty)\,$. To actually solve/factor it you could technically use the quartic formulas to get closed forms using radicals, but the result is not pretty. $\endgroup$ – dxiv Sep 17 '18 at 20:52
  • $\begingroup$ yes then thanks a lot. cant by basic manipulations be it possible? $\endgroup$ – maveric Sep 17 '18 at 20:53
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Let $$f(x)=x^8-14x^4-8x^3-x^2+1\implies f'(x)=8x^7-56x^3-24x^2-2x=0$$ for critical points. Then clearly $x=0$ is one, and is a maximum since $f''(0)<0$; $f(0)=1$.

We get $$g(x)=4x^6-28x^3-12x-1=0$$ and since $g$ is continuous, by the Mean Value Theorem, there is a minimum between $(-0.2,0)$ and a maximum between $(-0.4,-0.2)$. At these $x$, $f(x)$ is above the $x$-axis.

Hence there are at most four roots left. In a similar way, we test at $0.2$ intervals, and we find that $$f(-1.8)>0,\quad f(-1.6)<0\\f(-0.8)<0,\quad f(-0.6)>0\\f(0.2)>0,\quad f(0.4)<0\\f(2)<0,\quad f(2.2)>0.$$

Thus we have that $$(-1.6)^2+(-0.2)^2+0.2^2+2^2<r_1^2+r_2^2+r_3^2+r_4^2<(-1.8)^2+(-0.8)^2+0.4^2+2.2^2$$ or that $$6.64<r_1^2+r_2^2+r_3^2+r_4^2<8.88$$ so the only option must be $(C)$.

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  • $\begingroup$ but this is too much of approximation and working through options. wont there be any other way to find exact ans $\endgroup$ – maveric Sep 17 '18 at 19:14
  • $\begingroup$ Yes, @dxiv has given a much neater one. $\endgroup$ – TheSimpliFire Sep 17 '18 at 19:17
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Following the foundamental hint by dxiv, for $x^4-4x^2-x+1$ by Vieta's formula we have

  • $S_1=\sum r_i=-a_3=0$
  • $S_2=\sum r_ir_j=a_2=-4$
  • $S_3=\sum r_ir_jr_k=-a_1=1$
  • $S_4=r_1r_2r_3r_4=a_0=1$

and by Newton's sums we have that

  • $P_1=\sum r_i=S_1=0$
  • $P_2=\sum r_i^2=S_1P_1-2S_2=8$
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  • $\begingroup$ but you assuming first factor has all real roots!!! $\endgroup$ – maveric Sep 17 '18 at 20:43
  • $\begingroup$ yes that is seen. but A-B has all real roots !!how can we say that ? $\endgroup$ – maveric Sep 17 '18 at 20:47
  • $\begingroup$ second factor can be written as sum of squares so obviously no real root for it. $\endgroup$ – maveric Sep 17 '18 at 20:48
  • $\begingroup$ Yes you are right, I've overlooked that! It's a good point to clarify $\endgroup$ – user Sep 17 '18 at 20:53
  • $\begingroup$ @ashishdeosingh For the second factor we have that for $|x|\ge 1$ we have that $$x^4+4x^2+x+1\ge 5x^2+x+1 >0$$ and for $|x|<1$ we have $$x^4+4x^2+x+1\ge 5x^4+x+1>0$$ $\endgroup$ – user Sep 17 '18 at 20:59
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\begin{align} x^8-14x^4-8x^3-x^2+1&=0 \tag{1}\label{1} . \end{align}

\eqref{1} can be factored as

\begin{align} f_1(x)f_2(x)=0 ,\\ f_1(x)&=x^4+4x^2+x+1 \tag{2}\label{2} ,\\ f_2(x)&=x^4-4x^2-x+1 \tag{3}\label{3} . \end{align}

\begin{align} f_1(x)&=x^4+3x^2+(x^2+x+\frac14)-\tfrac14+1 \\ &=x^4+3x^2+\tfrac14(2x+1)^2+\tfrac34 >0\quad \forall x\in\mathbb{R} . \end{align}

We can also found that factor $f_2(x)$ has four distinct real roots, for example, observing that

\begin{align} f_2(-2)&=3>0 ,\\ f_2(-1)&=-1<0 ,\\ f_2(0)&=1>0 ,\\ f_2(1)&=-3<0 ,\\ f_2(3)&=43>0 , \end{align}

Now consider \begin{align} f_2(x)f_2(-x)&= (x^2)^4-8(x^2)^3+18(x^2)^2-9(x^2)+1 \tag{4}\label{4} . \end{align}

Expression \eqref{3} is a polynomial in $x^2$, its roots are squares of the roots of\eqref{1}, hence the sought sum of the squares of distinct real roots of \eqref{1} is $8$ (negated coefficient at $(x^2)^3$ in \eqref{4}).

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