4
$\begingroup$

let $k_{1}>k_{2}>k_{3}>k_{4}>k_{5}>k_{6}$ and $a=k_{1}+k_{2}+k_{3}+k_{4}+k_{5}+k_{6}$ and

$b = k_{1}k_{3}+k_{3}k_{5}+k_{5}k_{1}+k_{2}k_{4}+k_{2}k_{6}+k_{4}k_{6}$ and $c=k_{1}k_{3}k_{5}+k_{2}k_{4}k_{6}$

show that all the the roots of the equation $2x^3-ax^2+bx-c=0$ are real

want be able to go head, could some help me with this,thanks

  • 0
    The $k_i$ are real numbers? I am not sure what the better approach is, trying to calculate the roots and showing they are real or resorting to some ready criteria (e.g. positive discriminant). – 2017-01-09
  • 0
    Can you use calculus, say derivatives? – 2017-01-09
  • 0
    In your definition of $b$ perhaps $k_4k_6$ is mistyped as $k_2k_6$. – 2017-01-12

1 Answers 1

4

We can write $$2x^3-ax^2+bx-c=(x-k_1)(x-k_3)(x-k_5)+(x-k_2)(x-k_4)(x-k_6)$$ Let this be $f(x)$.

Then, we have $$f(k_1)\gt 0,\quad f(k_2)\lt 0$$ $$f(k_3)\lt 0,\quad f(k_4)\gt 0$$ $$f(k_5)\gt 0,\quad f(k_6)\lt 0$$

Since $f(x)$ is continuous, by the intermediate value theorem, there are three real numbers $\alpha,\beta,\gamma$ such that $$f(\alpha)=f(\beta)=f(\gamma)=0,\quad k_2\lt\alpha\lt k_1,\quad k_4\lt\beta\lt k_3,\quad k_6\lt\gamma\lt k_5$$

  • 1
    Great answer! Wonderful! – 2017-01-12