MHB What is the sum of all positive solutions?

[anemone]

Find the sum of the positive solutions to $5+x\lfloor x \rfloor-2x^2=0$.

 

[juantheron]

Spoiler

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]

[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]

 

[anemone]

Spoiler

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

Hi jacks, thanks for participating and I want to tell you that from the final set of solution that you have already figured it out, i.e. $\displaystyle \frac{\sqrt{21}-1}{2}<x \leq \sqrt{5}$, $\lfloor x \rfloor$ can be either 1 or 2. And from $\lfloor x \rfloor=1$, you will find the other remaining 2 solutions.:)

[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]

Well done Opalg and thanks for participating!