6
$\begingroup$

I need help with this inequality:

$\sqrt x +\sqrt{x+7} + 2\sqrt{x^2+7x} <35-2x$

It doesn't seem solvable. All roots of the corresponding equation are irrational.

$\endgroup$
  • 2
    $\begingroup$ What about $0 \leq x < \frac{841}{144}$ which the only root for the equation ? $\endgroup$ – Claude Leibovici Sep 10 '14 at 12:22
  • $\begingroup$ @ClaudeLeibovici The existence of a single (real) root doesn't guarantee that f(x) < 35 everywhere else. In the general case you need to show the function is positive for larger x , I think. $\endgroup$ – Carl Witthoft Sep 10 '14 at 17:17
  • $\begingroup$ @CarlWitthoft.You are perfectly correct, indeed. I was just mentionning that there is only one root which is rational. I wonder how this rationality could be established. Cheers :-) $\endgroup$ – Claude Leibovici Sep 10 '14 at 20:05
  • $\begingroup$ @ClaudeLeibovici well, once you've manipulated the inequality into an equivalent polynomial expression, there's some theorem about the number of real roots based on the number of coefficient sign changes. $\endgroup$ – Carl Witthoft Sep 11 '14 at 11:35
  • $\begingroup$ @CarlWitthoft. Thanks ! I learnt a lot from your answers and comments. Cheers :-) $\endgroup$ – Claude Leibovici Sep 11 '14 at 11:46
15
$\begingroup$

Let $a=\sqrt{x}+\sqrt{x+7}$, then $a+a^2<42$, and $-7<a<6$

$\endgroup$
  • $\begingroup$ Thanks. This is clever. I have problem with another one. Will post it later. $\endgroup$ – chen h. Sep 10 '14 at 11:57
  • $\begingroup$ It would appear, then, that the original problem was carefully constructed to allow this substitution. The cynic in me wonders whether this is a Kumon exercise. $\endgroup$ – Carl Witthoft Sep 10 '14 at 15:08
  • $\begingroup$ I'm just trying to refresh my math skills and I'm failing on how you get to your second step of a+a^2<42. Do you mind elaborating a touch? $\endgroup$ – Frank Shmrank Sep 10 '14 at 16:06
  • 1
    $\begingroup$ @DeanMacGregor $a^2 = x + 2\sqrt{x}*\sqrt{x+7} + (x+7)$ which reduces to $ 2x + 7 + 2\sqrt{x^2 +7*x} $ $\endgroup$ – Carl Witthoft Sep 10 '14 at 17:15
  • 1
    $\begingroup$ Note $\sqrt{x} + \sqrt{x + 7}$ is monotone in $x$, and that it is only defined for $x \geq 0$. So you need to solve $\sqrt{x} + \sqrt{x + 7} = 6$ and then the solution interval will be $[0,x]$. If you work it out, $x = ({29 \over 12})^2$, so the final answer is $[0, ({29 \over 12})^2]$. $\endgroup$ – Zarrax Sep 10 '14 at 17:36

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.