Help: Floor and Fractional Part Functions

How many real numbers between 1 to 23 satisfy $\lfloor x \rfloor \{ x \} = 2\sqrt x$ ?

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

That's a fun question. I think the best approach is to write $n=\lfloor x \rfloor$ , $t=\{x\}$ and put the equation in terms of $n,t$ . The idea is we know more about $n$ (it's an integer between $1$ and $23$ ) and $t$ (it's non-negative real less than $1$ ) than we do about $x$ alone.

So $\begin{aligned} nt &= 2\sqrt{n+t} \\ n^2 t^2 &= 4n + 4t \\ n^2 t^2 - 4t-4n&=0 \end{aligned}$

This last equation is quadratic in $t$ ; so we know it has to have a real root somewhere in $0\le t<1$ .

Is that enough for you to finish this off? (By the way, I make the answer $18$ )

Chris Lewis - 3 months, 3 weeks ago

It's enough for me to finish mate, thanks. I used quadratic formula(solving for $t$ ) and using the inequality $0≤t<1$ and $n=[1,22]$ .

$\frac{4+\sqrt{16+16n³}}{2n²}<1$

$4+4\sqrt{n³+1}<2n²$

$2\sqrt{n³+1}<n²-2$

$4n³+4<n^{4}-4n²+4$

$n^{4}-4n³-4n²>0$

$n²(n²-4n-4)>0$

$n²-4n-4>0$

$n²-4n+4>8$

$(n-2)²>8$

$n>2+2\sqrt{2}$

Meaning $n=[5,22]$ , so we have $18$ real numbers. But I'm not sure if this is the best approach, not confident with it. lol

Ryan Merino - 3 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$