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?

  • $\begingroup$ 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

Hint:   the real roots are the roots of the first factor:

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

| cite | improve this answer | |
  • $\begingroup$ 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

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 $(C)$.

| cite | improve this answer | |
  • $\begingroup$ 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

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

  • $P_1=\sum r_i=S_1=0$
  • $P_2=\sum r_i^2=S_1P_1-2S_2=8$
| cite | improve this answer | |
  • $\begingroup$ 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

\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}).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.