2

The product of two of the roots of $$x^4 -11x^3 + kx^2 + 269x - 2001=0$$ is $-69$. Find k.

This is a question I have recently received, and I am required to take a test on related questions tomorrow. I have absolutely no idea on how to start with this or even which concept this belongs to. I would greatly appreciate it if someone tells me a link online to learn the theory required to answer this question and a page where I can learn how to solve questions similar to this. A solution would also be appreciated. I am only in 9th grade so I am absolutely clueless. Apologies for the amateurish question.

psmears
  • 767
Shreyash Chaudhari
  • 147
  • 2
    Isn't there an essential part of speach missing from your question? Should it be something like "The product of two of the roots of is "? – Hagen von Eitzen Dec 27 '14 at 13:30
  • 2
    Check Vietas formulas on wikipedia http://en.wikipedia.org/wiki/Vieta%27s_formulas. The quadratic and cubic case have been worked out, but you could use the quartic case. – Myself Dec 27 '14 at 13:30

2 Answers2

9

Suppose the roots are $a,b,c$ and $d.$ Then by Vietas formulas, $$a+b+c+d=11\to(1)$$ $$ab+ac+ad+bc+bd+cd=k\to(2)$$ $$abc+abd+acd+bcd=-269\to(3)$$ $$abcd=-2001\to(4).$$ By your additional condition (suppose) $$ab=-69.$$ Then by $(4)$$$cd=29$$ and by $(3)$ $$ab(c+d)+cd(a+b)=-69(c+d)+29(a+b)=-269.$$ $$-69(c+d)+29(11-(c+d))=-269.$$ Hence $$c+d=6$$ and $$a+b=5.$$ Finally $$ab+ac+ad+bc+bd+cd=ab+(a+b)(c+d)+cd=k.$$ Hence $k=-10.$

Bumblebee
  • 18,220
  • 5
  • 47
  • 87
  • 1
    It might be mentioned that the last equation gives $k=-10$. You have essentially factored the polynomial as $(x^2-5x-69)(x^2-6x+29)$. – Marc van Leeuwen Jan 02 '15 at 11:40
2

The following is basically the answer by Nilan, reformulated to avoid explicit references to each of the $4$ roots (since in any case one cannot distinguish the two whose product is given form each other, nor the remaining two from each other).

Your polynomial must factor as product of two monic quadratic polynomials, the first one with the two roots whose product is $-69$. This gives an equation $$ x^4 -11x^3 + kx^2 + 269x - 2001 = (x^2 + px -69)(x^2+qx+r) $$ where $p,q,r$ are to be determined. Comparing coefficients gives the equations $$ \begin{align} -11 &= p+q, \\ k &= r+pq-69, \\ 269 &= pr - 69 q, \\ -2001 &= -69r. \end{align} $$ The final equation gives $r=29$. Then the first and third equations become a linear system in $p,q$, whose solution is $p=-5,q=-6$. Finally the second equation gives $k=29+30-69=-10$.

Marc van Leeuwen
  • 115,048