1

What is the number of real roots of $$x^6-3x^2+1=0$$ ?

I know that there are $6$ roots for this polynomial as the highest power is $6$ but how we could determine number of real roots or complex roots or repeated roots ?

Thank you for your help

user373141
  • 2,503
  • 3
    Let $a$ be a real root of this polynomial. Then $a^2$ is a (positive!) root of $x^3-3x^2+1$. How many positive roots does $x^3-3x^2+1$ have? – Crostul Apr 11 '17 at 08:14
  • How about trying Wolfram Alpha? That engine answers questions like this. – Eugene Apr 11 '17 at 08:26
  • 1
    @Eugene: right, but this is the best way to not learn. –  Apr 11 '17 at 08:27

3 Answers3

2

If you introduce a new variable $y=x^2$, you get an equation with a polynomial of degree $3$.

This polynomial has $3$ zeroes, all real, one negative and two positive (because it is negative at $-3$ positive at $0$, negative at $1$ and positive at $2$).

Each of the two positive zeroes corresponds to two zeroes of the original polynomial.

5xum
  • 123,496
  • 6
  • 128
  • 204
1

Let $t=x^2\ge0$ and $P(t):=t^3-3t+1$.

Then $P'(t)=3t^2-3$ cancels for $t=\pm1$, which correspond to the extrema of $P$, of which only $(1,-1)$ matters to us.

As $P(0)>0$, $P(1)<0$ and $P(\infty)=\infty$, we have two positive roots in $t$, hence four roots in $x$.

This is well confirmed by a plot.

enter image description here

0

Let $f(x)=x^3-3x+1$.

Easy to see that $f(-3)<0$, $f(0)>0$,$f(1)<0$ and $f(2)>0$,

which says that the equation $x^3-3x+1=0$ has two positive roots

and from here your equation has four roots.

If $x^2=2\cos\alpha$ for starting equation than we get $\cos3\alpha=-\frac{1}{2}$ and from here you can get four roots.

Michael Rozenberg
  • 194,933
  • why did you choose x={-3,0,1,2} ; Is it arbitrary ? – user373141 Apr 11 '17 at 08:33
  • 1
    @prayersmith I used the same values, I got them after I checked the integer values from $-5$ to $5$. You could also look at the plot of $f$ generated by wolfram alpha to get a better picture. – 5xum Apr 11 '17 at 08:40