0
$\begingroup$

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?

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ While you're at it, showing that the second derivative is negative makes things easier than stopping at the first derivative. And it is concave, not convex. $\endgroup$ – Julien May 4 '13 at 13:37
  • $\begingroup$ @julien $f'$ can be direct seen being decrease, so there is no need to go to second derivative. $\endgroup$ – Ma Ming May 4 '13 at 13:40
  • $\begingroup$ 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. $\endgroup$ – Julien May 4 '13 at 13:43
  • $\begingroup$ @julien You need to calculate the second derivative. $\endgroup$ – Ma Ming May 4 '13 at 13:46
  • $\begingroup$ If you are to write this down, it is way more cumbersome to justify that the derivative decreases (it takes at least two sentences), than to simply compute the second derivative which is trivially negative. $\endgroup$ – Julien May 4 '13 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.