MHB What is the sum of all positive solutions?

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Positive Sum
AI Thread Summary
The equation to solve is \(5 + x\lfloor x \rfloor - 2x^2 = 0\), leading to a quadratic form where \(a = \lfloor x \rfloor\). The positive solutions are derived from \(x = \frac{1}{4}(a + \sqrt{a^2 + 40})\), requiring \( \lfloor x \rfloor = a\). Valid solutions occur for \(a = 1\) and \(a = 2\), yielding positive solutions \(x = \frac{1}{4}(1 + \sqrt{41})\) and \(x = \frac{1}{4}(2 + \sqrt{44})\). The sum of these solutions is \(\frac{3}{4} + \frac{1}{4}(\sqrt{41} + \sqrt{44})\).
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Find the sum of the positive solutions to $5+x\lfloor x \rfloor-2x^2=0$.
 
Mathematics news on Phys.org
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
 
[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]
 
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

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.:)

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]

Well done Opalg and thanks for participating!:cool:
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top