12
$\begingroup$

If $x^6-12x^5+ax^4+bx^3+cx^2+dx+64=0$ has positive roots then find $a,b,c,d$.

I did something but that don't deserve to be added here, but what I thought before doing that is following:

  1. For us, Product and Sum of roots are given.
  2. Roots are positive.
  3. Hence I tried to use AM-GM-HM inequalities, as sum and product are given, but I failed to conclude something good.

So please deliver some hints or solution regarding AM-GM-HM inequalities.

$\endgroup$

2 Answers 2

16
$\begingroup$

Use AM-GM on the roots.

Say $a_1,a_2,a_3,a_4,a_5,a_6$ are the roots of the equation.

Then $a_1+a_2+a_3+a_4+a_5+a_6\geq 6(a_1\times a_2\times a_3\times a_4\times a_5\times a_6)^{1/6} $

Now $a_1+a_2+a_3+a_4+a_5+a_6=12$

And also $6(a_1\times a_2\times a_3\times a_4\times a_5\times a_6)^{1/6}=12 $

Hence equality condition of AM-GM inequality holds, which implies that all roots are equal, and you know the sum of the roots is $12$, hence all roots are equal to $2$, now find $a,b,c,d$.

$\endgroup$
1
  • $\begingroup$ eh! it feels so bad giving up on the problem after approaching so close to it, between thank you. $\endgroup$
    – mnulb
    Commented Jun 2, 2017 at 2:09
6
$\begingroup$

Let $x_i$ be our roots.

Now, by AM-GM $$2=\frac{\sum\limits_{i=1}^6x_i}{6}\geq\sqrt[6]{\prod_{i=1}^6x_i}=2,$$ which says that all $x_i=2$.

Thus, $$a=2^2\cdot\frac{6(6-1)}{2}=60,$$ $$b=-2^3\cdot\frac{6(6-1)(6-2)}{6}=-160,$$ $$c=2^4\cdot\frac{6(6-1)(6-2)(6-3)}{24}=240$$ and $$d=-2^5\cdot\frac{6(6-1)(6-2)(6-3)(6-4)}{120}=-192.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .