How to Prove monotony of a Irrational Function

Question

I have the following equation:

$x+\sqrt{3x+7}+\sqrt{5x+14} = \sqrt 7 + \sqrt{14}$

The obvious solution is 0 but how can I prove that only 0 is the right answer and there aren't any other roots. Thank you in advance

Answer 1

I suppose calculus is not even necessary to show the square root function is increasing.

Assume $0\leq x<y$. Then it must be the case that $\sqrt{x}<\sqrt{y}$. Otherwise, we would have $\sqrt{y} \leq \sqrt{x}$, so $y= \sqrt{y}\cdot \sqrt{y} \leq \sqrt{x}\cdot \sqrt{x} = x$. But $y \leq x$ cannot be true since we assumed $x<y$. Hence the square root function is increasing.

Now let's use this to show that $x=0$ is the only solution to the equation you wrote. If $y<x=0$, then $3y+7<3x+7$ and $5y+14<5x+14$. Since we showed the square root function is increasing, we have $\sqrt{3y+7}<\sqrt{3x+7}$ and $\sqrt{5y+14}<\sqrt{5x+14}$. Combining all of these inequalities, $$y+\sqrt{3y+7}+\sqrt{5y+14}<\sqrt{3x+7}+\sqrt{5x+14} = \sqrt{7}+\sqrt{14}.$$ So $y$ cannot be a solution. You should be able to write a similar argument for the case $y>x=0$ to show that $y$ cannot be a solution either.

Answer 2

Sorry, don’t have a computer to type solution.