# Show that $x^{35}+\dfrac{20205}{2+x^{17}+\cos^2x}=100$ has no root $x\in \mathbb{R}$

Show that $x^{35}+\dfrac{20205}{2+x^{17}+\cos^2x}=100$ has no root $x\in \mathbb{R}$.

By plotting graph I have seen that there are no roots for $x$. Can somebody prove it theoretically?

-
+1 Nice question. Can you tell us the source of this question? – Srivatsan Oct 8 '11 at 11:14
@SrivatsanNarayanan It's given in one my homework assignments. – Ramana Venkata Oct 8 '11 at 11:15
I've retagged away from (theory-of-equations): the more common connotations of that phrase considers algebraic equations. The fact that you have a $\cos$ factor in there makes your equation transcendental, and outside what would traditionally be considered as theory of equations. – Willie Wong Oct 17 '11 at 15:22
For what it is worth, for actual questions on theory of equations, you may considered instead the (galois-theory) and (field-theory) tags as appropriate. – Willie Wong Oct 17 '11 at 15:24

Here's an answer, but I am sure there should be better ones.

Assume $x \in \mathbb R$ to be a root of the given equation. Rearrange the equation as $$(x^{35}-100) (x^{17}+2+\cos^2 x) = - 20205.$$ Let $A := x^{35}-100$ and $B := x^{17}+2+\cos^2 x$. Since $AB < 0$, exactly one of $A$ and $B$ is positive and the other is negative.

If $B < 0$ and $A > 0$, then $x^{35} > 100 > 0$, which implies that $x > 0$. Then we can conclude that $B = x^{17}+2 + \cos^2 x > 0$, which contradicts the assumption about the sign of $B$. So this case is impossible.

We are left with the case $A < 0$ and $B > 0$.

• From $A < 0$, we get $x^{35} < 100$. Therefore $x^{17} < 10$. Hence, $B = x^{17}+2+\cos^2 x \leq x^{17}+3 < 13$.

• Similarly, from $B > 0$, we get $x^{17} > -2 - \cos^2 x \geq -3$. Therefore, $A = x^{35} - 100 > (-3)^{3}-100 = -127$. In other words, $|A| < 127$.

Multiplying the upper bounds on $|A|$ and $B$, we get $$|A \cdot B| = |A| \cdot B < 127 \cdot 13 = 1651 ,$$ which contradicts the equation $AB = - 20205$ we started out with.

-
I was also actually trying to solve the question in two intervals $x>0$ and $x<0$. I was struck in $x<0$ case. Great Solution. Thanks. – Ramana Venkata Oct 8 '11 at 11:59
@Ramana Yes, somehow splitting it into $x > 0$ and $x < 0$ does not work by itself. I guess (also see Ragib's answer) further splitting each of those two cases into subcases depending on the size of $x$ works. – Srivatsan Oct 8 '11 at 12:03