0
$\begingroup$

For example, to prove that the function $x^8+x-1=0$ has exactly two roots, we first prove that $f$ has at most $2$ real roots by using differentiation. $f'(x)=8x^7+1=0$, by using that the function has two roots and therefore it should have at least one point such that $f '(c)=0$ and then we use IVT to prove that it has exactly two roots. First of all, Are these correct? If yes, when I'm given a function such as $5x^4-9x^2-1=g(x)$ to prove that it has exactly $2$ roots, because its derivative has $3$ roots, I cannot first prove that it has at most $2$ roots and cannot solve it.

So as you see, my question is how we can prove that $5x^4-9x^2-1=0$ has exactly two roots.

$\endgroup$
  • 1
    $\begingroup$ $g(x)=h(x^2)$ where $h(x)=5x^2-9x-1$. Note that $h(x)=0$ has two solutions, only one of them non-negative. The argument for $x^8+x-1$ is OK. $\endgroup$ – conditionalMethod Dec 6 '19 at 21:26
  • $\begingroup$ Okay, but what should I do if I am given x^4-6x^2-8x+1=0, its derivative has 2 roots but still I should try to prove that it has exactly two roots. $\endgroup$ – Auroras Dec 6 '19 at 21:30
  • $\begingroup$ Welcome to MSE! If you look at the edit. You can see how the mathematical type was made using \$ signs. Can you make sure to do this when you ask questions here: it makes it a little easier on the eyes. $\endgroup$ – Mason Dec 6 '19 at 21:32
  • $\begingroup$ Well, the problem of counting real roots can be solved algorithmically using Sturm's theorem. It is not the most efficient, but at least it is short to explain and read. $\endgroup$ – conditionalMethod Dec 6 '19 at 21:33
  • $\begingroup$ You can also weed out many problems with little work if you can apply Descartes' rule of signs, which for example applied to $x^8+x-1$ tells you that it has exactly one positive root, and changing $x$ to $-x$ and applying it to $x^8-x-1$ tells it has exactly one negative root. $\endgroup$ – conditionalMethod Dec 6 '19 at 21:50
1
$\begingroup$

Welcome!

It's purely algebraic, and uses only tools from high school:

This is a biquadratic equation, so set $t=x^2$. It is a nonnegative root of the quadratic equation: $$5t^2-9t-1=0.$$ Now this quadratic has a positive discriminant, so it has two roots. However, as the leading coefficient and the constant term have opposite signs, Vieta's relations imply that root is positive and the other is negative.

Can you conclude now?

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Let $f(x)=5x^4-9x^2-1$,$f'(x)=20x^3-18x$ has three roots from which two of them are opposite.This means critical numbers are 0 and $\sqrt{.9}$ and$-\sqrt{.9}$ This means the graph of the function has two minimm and a max at $x=0$. since $f(c)<0$ "c is a none zero critical number", the result follows.

| cite | improve this answer | |
$\endgroup$

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.