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anemone
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Find the sum of the positive solutions to $5+x\lfloor x \rfloor-2x^2=0$.
jacks said:Using The formula $(x-1)<\lfloor x \rfloor \leq x$
So $x^2-x < x\lfloor x \rfloor \leq x^2\Rightarrow x^2-x+5 < x\lfloor x \rfloor +5 \leq x^2$
So $x^2-x+5 -2x^2 < x\lfloor x \rfloor +5-2x^2 \leq x^2+5-2x^2$
So $-x^2-x+5 < 5+x\lfloor x \rfloor -2x^2 \leq -x^2+5$
So $-x^2-x+5 < 0 \leq -x^2+5$
Now Solving $-x^2-x+5 < 0$, we get $\displaystyle x> \frac{\left(\sqrt{21}-1\right)}{2}\cup x< \frac{\left(\sqrt{-21}-1\right)}{2}$
Now Solving $-x^2+5\geq 0$ , we get $\sqrt{5}\leq x \leq \sqrt{5}$
Now Solution of $-x^2-x+5 < 0 \leq -x^2+5$ is $\displaystyle \frac{\sqrt{21}-1}{2}<x \leq \sqrt{5}$
So We get $\lfloor x \rfloor = 2$ means $2\leq x<3$
Now given $5+x\lfloor x \rfloor -2x^2 = 0$ put $\lfloor x \rfloor = 2\;,$ we get
$5+2x-2x^2 = 0$ ,we get $\displaystyle x = \frac{\sqrt{11}+1}{2}$
But I have missed $2$ more solution.
I did not understand where i have missed.
Thanks
Opalg said:[sp]Let $a = \lfloor x \rfloor$. The solutions of the quadratic equation $5+ax-2x^2=0$ are $\frac14\bigl(a \pm\sqrt{a^2+40}\bigr)$, and to get a positive solution we must take the positive square root. So we want to find all the non-negative integers $a$ for which $\left\lfloor \frac14\bigl(a + \sqrt{a^2+40}\bigr) \right\rfloor = a$.
When $a=0$, the solution to the quadratic is $x = \frac14\sqrt{40} \approx 1.58$ and $\lfloor x \rfloor = 1$, which is too big.
When $a=1$, the solution to the quadratic is $x = \frac14\bigl(1 + \sqrt{41}\bigr) \approx 1.85$ and $\lfloor x \rfloor = 1$. So $x = \frac14\bigl(1 +\sqrt{41}\bigr)$ qualifies as a solution.
When $a=2$, the solution to the quadratic is $x = \frac14\bigl(2 + \sqrt{44}\bigr) \approx 2.16$ and $\lfloor x \rfloor = 2$. So $x = \frac14\bigl(2 +\sqrt{44}\bigr)$ qualifies as a solution.
When $a\geqslant3$, $\sqrt{a^2+40} \leqslant a+4$ and so $\frac14\bigl(a + \sqrt{a^2+40}\bigr) \leqslant \frac12a+1$, which is less than $a$ and therefore too small.
So the only solutions occur when $a=1$ or $2$, and the sum of those solutions is $\frac34 + \frac14\bigl(\sqrt{41} + \sqrt{44}\bigr).$[/sp]
The sum of all positive solutions refers to the total value obtained when adding together all of the solutions to a given problem or equation that are greater than zero.
The sum of all positive solutions can be found by first solving the problem or equation to obtain all of the solutions, and then adding together only the positive solutions.
No, the sum of all positive solutions cannot be negative because by definition, positive solutions are greater than zero and therefore cannot result in a negative sum.
No, it is not necessary to include negative solutions when finding the sum of all positive solutions. In fact, including negative solutions would result in a sum that is not solely representative of the positive solutions.
Yes, the sum of all positive solutions can be infinity if there are an infinite number of positive solutions to the given problem or equation. This is often the case with infinite series or sequences.