Disproving existence of real root in some interval for a quintic equation Disprove the statement: There is a real root of equation $\frac{1}{5}x^5+\frac{2}{3}x^3+2x=0$ on the interval (1,2).
I am not sure whether to prove by counter-example or by assuming the statement is true and then proving by contradiction. 
 A: First step: what is the value of the polynomial for $x=1$?
Second step: what is the sign of the derivative of this polynomial on the interval $[1,2]$? 
Third step: conclusion.
edit
Altenative: $x\ge 1$, therefore $x^n\ge 1$ for all $n\ge 1$. Hence $$\frac{1}{5}x^5+\frac{2}{3}x^3+2x\ge \frac{1}{5} +\frac{2}{3} +2 $$
whenever $x\ge 1$. 
A: All terms are increasing on $\Bbb R$, so the sum is also increasing. There is the trivial root $0$, hence it's the only root.

Alternately, you can factor out $x$, then you are left with a biquadratic equation $\frac15x^4+\frac23x^2+2=0$, then let $x^2=y$, and you have a trinomial to solve. But you don't even have to: it's obviously always positive.
A: Hint: $x$ can't be a root on the interval, as it is positive throughout, so you can divide through by $x$ and solve the resulting quadratic in $x^2$
A: Odd-degree polynom with real coefficients always has at least one real root, as all complex roots of such polynom are conjugate pairs.
A: It's sufficient to show that:


*

*f(1) > 0

*For all n > 0: f(1 + n) > f(1) 
Step #2 means that the function is increasing at all points to the right of x=1.  From this you can say that f(x) doesn't equal zero (has a real root) in the range (1,2) because f(x) is always above 1 in that range.
Working out the algebra:
f(1 + n) = (1/5 + 2/3 + 2) + 4n + 4n^2 + (2 + 2/3)n^3 + n^4 + (1/5)n^5

Then to show that f(1 + n) > f(1), you have to show: 
(1/5 + 2/3 + 2) + 4n + 4n^2 + (2 + 2/3)n^3 + n^4 + (1/5)n^5 > (1/5 + 2/3 + 2)

Canceling the (1/5 + 2/3 + 2) gives:
4n + 4n^2 + (2 + 2/3)n^3 + n^4 + (1/5)n^5 > 0

It should be easy to see that as n increases, this inequality remains true, as long as n > 1.
Edit: here's one way to work this out:
Divide: by 4: n + n^2 + (3/4)n^3 + (1/4)n^4 + (1/20)n^5 > 0
We know that n>1 so we can add a second inequality which removes the initial n +:
n + n^2 + (3/4)n^3 + (1/4)n^4 + (1/20)n^5 > n^2 + (3/4)n^3 + (1/4)n^4 + (1/20)n^5 > 0

Subtract out the middle pieces to make the middle zero:
n > 0 > -n^2 - (3/4)n^3 - (1/4)n^4 - (1/20)n^5

A: Factor $x ( 1/5 x^4 + 2/3 x^2 + 2 )$.  So one root is $x=0$.
Let $z = x^2$, then we have $1/5 z^2 + 2/3 z + 2 = 0$.  The discriminant1 is $(4/9-8/5)$, which is negative, so there are no real roots in $z$.  A real root in $x$ implies a real root in $z$, so $x = 0$ is the only real root in $x$.

1  The term $(b^2 - 4 ac)$ within the square root portion of the quadratic equation.
