i need to use with Rolle's theorem to prove that this Polynomial has no then 2 Square root, Regardless of the values m and n.

i not find any Square root is can be?

someone can show me the correct way to Solve it ?

thanks.

Printable View

- Dec 9th 2010, 12:10 AMZOOZprove the Square root
i need to use with Rolle's theorem to prove that this Polynomial has no then 2 Square root, Regardless of the values m and n.

i not find any Square root is can be?

someone can show me the correct way to Solve it ?

thanks. - Dec 9th 2010, 01:33 AMHallsofIvy
This problem, as stated, does not require you to

**find**any roots, just determine that roots exist. (And not "square roots". Square roots are specifically roots to the equation $\displaystyle x^2= a^2$.) It's not clear to me what you mean by "no then 2 roots". What does the "then" mean? Do you mean that, for all m and n, it has either 0 or 2 roots, but never 1, 3, or 4?

Rolle's theorem says that if f(a)= f(b)= 0 then there exist some number, c, between a and b, such that f'(c)= 0. You want to show that $\displaystyle x^4- 2x^3+ 6x^2+ mx+ n= 0$. If you take that to be f', then $\displaystyle f(x)= \frac{1}{5}x^5- \frac{1}{2}x^4+ 2x^3+ \frac{m}{2}x^2+ mx+ C$ for some number C. You can always choose C to make f= 0 for a given x. - Dec 9th 2010, 01:48 AMBAdhi
functions are not printed well sir,

$\displaystyle x^4- 2x^3+ 6x^2+ mx+ n= 0$

$\displaystyle f(x)= \frac{1}{5}x^5- \frac{1}{2}x^4+ 2x^3+ \frac{m}{2}x^2+ mx+ C$ - Dec 9th 2010, 04:23 AMZOOZ
well yes i mean than 2 root.

but i not Understand what you Write: "If you take that to be f', then [tex]f(x)= \frac{1}{5}x^5- \frac{1}{2}x^4+ 2x^3+ \frac{m}{2}x^2+ mx+ C[tex]: what is mean?

did you mean f'=4x^3+6x^2+12x+m?

thanks - Dec 9th 2010, 04:24 AMZOOZ
how did you get it ?