2
$\begingroup$

If$\frac{1+\alpha}{1-\alpha},\frac{1+\beta}{1-\beta},\frac{1+\gamma}{1-\gamma}$ are the roots of the cubic equation $f(x)=0$ where $\alpha,\beta,\gamma$ are the real roots of the cubic equation $3x^3-2x+5=0$,then find the number of negative real roots of the equation $f(x)=0$.

My attempt:I tried finding out solution of $3x^3-2x+5=0$ to get $\alpha,\beta,\gamma$ by rational root method and hit and trial method but could not get them.Is my approach correct? Or Descretes rule is to be applied?

Can someone help me solve this question?

$\endgroup$
3
  • $\begingroup$ So what you are asking is if $\alpha,\beta,\gamma$ are the real roots of $3x^2-2x+5=0$, then how many of $(1+\alpha)/(1-\alpha),(1+\beta)/(1-\beta),(1+\gamma)/(1-\gamma)$ are negative? (I don't see where $f$ comes in.) $\endgroup$
    – ajd
    Jul 28, 2015 at 6:47
  • $\begingroup$ Isn't this just convoluted for: How many of $\frac{1+\alpha}{1-\alpha}$, $\frac{1+\beta}{1-\beta}$, $\frac{1+\gamma}{1-\gamma}$ are negative? $\endgroup$ Jul 28, 2015 at 6:47
  • 1
    $\begingroup$ You said $\alpha, \beta, \gamma$ are the real roots of the cubic, but the cubic has 2 complex solutions... $\endgroup$
    – user217285
    Jul 28, 2015 at 6:52

1 Answer 1

3
$\begingroup$

Note that $\frac{1+\alpha}{1-\alpha}<0$ is equivalent to $1+\alpha$ and $1-\alpha$ having opposite signs, which happens when $\alpha>1$ or $\alpha <-1$. So all we have to do is count the number of roots of $3x^2-2x+5$ in $[-1,1]$.

Let $p(x) = 3x^3-2x+5$. Now $p(-1)=4$ and $p(1)=6$. Moreover, $p'(x)=9x^2-2$, which has roots at $x=\pm\sqrt{2}/3$. Also, $p''(x)=18x$, which has the sign the same as $x$, so $x=-\sqrt{2}/3$ is a local maximum of $p$ while $x=\sqrt{2}/3$ is a local minimum of $p$. In particular, the minimum of $p$ on $[-1,1]$ is the minimum of $p(-1)=4$, $p(1)=6$, and $p(\sqrt{2}/3)$. Now $p(\sqrt{2}/3)=3(\sqrt{2}/3)^3-2\sqrt{2}/3+5=-(4/9)\sqrt{2}+5>0$. Therefore, $p$ has no roots on $[-1,1]$, so the answer to your question is simply the number of real roots of $p$, which is $1$.

$\endgroup$

You must log in to answer this question.

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