# How to find the number of real roots of a polynomial

no of real roots of the equation $(97-x)^{1/4} + x^{1/4}=5$

The options for the amount of real amounts-$1$ real root, $2$ real root, $3$ real root, $4$ real root.

I got the answer as $2$ real roots being $x=16,81$. Is it correct? Is there any general method for finding roots in these cases?

• What are the options? It seems you just put 1,2,3, and didn't list what they actually were. – Jeel Shah May 4 '13 at 13:22
• those are the options-1 real root, 2 real roots, 3 real roots or 4 real roots. – kris91 May 4 '13 at 13:23
• First, cheat. Then, explain. – Julien May 4 '13 at 13:27
• Is it normal to call this a polynomial? – Karl Kronenfeld May 4 '13 at 13:30

Let $97-x=a^4$ and $x=b^4$ where $a,b\in\Bbb R$, then $$a^4+b^4=97$$ and $$a+b=5$$ which then gives $$a^4+(5-a)^4=97$$ which is a quartic equation and it's solution is given here
Hint: Let $f(x)=x^{1/4}+(97-x)^{1/4}-5$. Consider its derivative, you will find $f'$ is decrease, so $f$ is convex upward.
• @julien $f'$ can be direct seen being decrease, so there is no need to go to second derivative. – Ma Ming May 4 '13 at 13:40
• It is trivial that $f''<0$. Is is easy that $f'$ decreases. Trivial is easier than easy. What's wrong about taking the second derivative. Hessian, etc...when one has a second derivative, it is always preferable to use it when it comes to convexity. – Julien May 4 '13 at 13:43