# prove the Square root

• Dec 9th 2010, 01:10 AM
ZOOZ
prove 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, 02:33 AM
HallsofIvy
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 $x^2= a^2$.) 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 $x^4- 2x^3+ 6x^2+ mx+ n= 0$. If you take that to be f', then $f(x)= \frac{1}{5}x^5- \frac{1}{2}x^4+ 2x^3+ \frac{m}{2}x^2+ mx+ C$ for some number C. You can always choose C to make f= 0 for a given x.
• Dec 9th 2010, 02:48 AM
functions are not printed well sir,

$x^4- 2x^3+ 6x^2+ mx+ n= 0$
$f(x)= \frac{1}{5}x^5- \frac{1}{2}x^4+ 2x^3+ \frac{m}{2}x^2+ mx+ C$
• Dec 9th 2010, 05:23 AM
ZOOZ
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, 05:24 AM
ZOOZ
how did you get it ?