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.

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    $\begingroup$ What about $0 \leq x < \frac{841}{144}$ which the only root for the equation ? $\endgroup$ Sep 10, 2014 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. $\endgroup$ Sep 10, 2014 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 :-) $\endgroup$ Sep 10, 2014 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. $\endgroup$ Sep 11, 2014 at 11:35
  • $\begingroup$ @CarlWitthoft. Thanks ! I learnt a lot from your answers and comments. Cheers :-) $\endgroup$ Sep 11, 2014 at 11:46

1 Answer 1


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

  • $\begingroup$ Thanks. This is clever. I have problem with another one. Will post it later. $\endgroup$
    – chen h.
    Sep 10, 2014 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. $\endgroup$ Sep 10, 2014 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? $\endgroup$ Sep 10, 2014 at 16:06
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    $\begingroup$ @DeanMacGregor $a^2 = x + 2\sqrt{x}*\sqrt{x+7} + (x+7)$ which reduces to $ 2x + 7 + 2\sqrt{x^2 +7*x} $ $\endgroup$ Sep 10, 2014 at 17:15
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    $\begingroup$ 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, 2014 at 17:36

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