0

find all real values of $x$ in $x^4-2x^3+8x-3=0$

what i try

$$x^4-2x^3+x^2=x^2-8x+3$$

$$x^2(x-1)^2=x^2-8x+3$$

Did not know how to find $x$. Help me please

jacky
  • 5,345

1 Answers1

3

For any $k$ we have $$x^4-2x^3+8x-3=(x^2-x+k)^2-2kx^2+2kx-x^2-k^2+8x-3=$$ $$=(x^2-x+k)^2-((2k+1)x^2-2(k+4)x+k^2+3).$$ Now, choose a value of $k$ such that $$2k+1>0$$ and $$(k+4)^2-(2k+1)(k^2+3)=0.$$ After this you'll get two quadratic equations.

Can you end it now?

Michael Rozenberg
  • 203,855
  • 2
    The method of Ferrari. The problem is, you have to solve the resolvent cubic first. Also, can you guarantee that this cubic has a root in the interval $(-1/2,+\infty),$ as you stipulate? – Allawonder Feb 22 '20 at 07:58
  • 1
    Ferrari's method. – Jyrki Lahtonen Feb 22 '20 at 07:59
  • @Allawonder We can prove that if the resolvent has three real roots, so one of them we'll give a difference of squares. Wile if the resolvent has unique real root, so this root will give a difference of squares. Id est, this method always gives a difference of squares and always works. By the way, our resolvent has unique real root, which we can get by the Cardano's formula. – Michael Rozenberg Feb 22 '20 at 16:17
  • 1
    @MichaelRozenberg I know that a cubic always has a real root. My question is whether you can in this case get a root in the interval $(-1/2,+\infty),$ which is what you require in order for the quartic to have only real roots. – Allawonder Feb 22 '20 at 16:37
  • @Allawonder See please better my previous comment. We always can get a value of $k$ from the interval $\left[-\frac{1}{2},+\infty\right).$ – Michael Rozenberg Feb 22 '20 at 16:59
  • @MichaelRozenberg How do you know that the resolvent has a root in this interval? – Allawonder Feb 22 '20 at 17:18
  • @Allawonder In the general? For our equation the resolvent gives the equation $2k^3-2k-13=0,$ which has unique real root. Now, you can read better my preprevious comment and this root must be in $\left[-\frac{1}{2},+\infty\right)$ or you can just to calculate this root by the Cordano's formula, which gives the same result. – Michael Rozenberg Feb 22 '20 at 17:27
  • @MichaelRozenberg I know that this has a real root. My question is whether the root lies in the interval. You've not yet demonstrated that there's a root in the required interval. – Allawonder Feb 22 '20 at 17:30
  • @Allawonder The polynomial $2k^3-2k-13$ has a positive root $k_0$. Do you see it? If so, $2k_0+1>0.$ – Michael Rozenberg Feb 22 '20 at 17:36
  • 1
    @MichaelRozenberg The question is, does OP see it? Shouldn't you include that in your answer as well? That's the point I've been trying to make -- that you should at least mention that the inequality and the equation in $k$ are consistent. – Allawonder Feb 22 '20 at 17:38
  • @Allawonder My post was a hint only. It was accepted, which says that a topic starter saw it. Also, see please better my first comment. There I said what happens in the general case. – Michael Rozenberg Feb 22 '20 at 17:41