Real Solutions for 4x^2-40x+51=0 and 4x^2-40[x]+51=0

[anemone]

Find all real solutions for the system \(\displaystyle 4x^2-40\left\lfloor{x}\right\rfloor+51=0\).

 

[kaliprasad]

Find all real solutions for the system \(\displaystyle 4x^2-40\left\lfloor{x}\right\rfloor+51=0\).

Spoiler

First $4x^2+ 51$ is positive. So x has to be positive else the sum is positive
we have $4x^2 - 40 \lfloor x \rfloor + 51 = 0$ => $4x^2 - 40 x + 51 <= 0$ equal only when x is integer and $4(x+1)^2 - 40 x + 51 >0$
$4x^2- 40 x + 51 < 0$
$=> 2(x-5)^2 < 49$
or $=> (x-5) < 3.5$
$=> x < 8.5$
$4(x+1)^2 - 40 x + 51 >0$
$=>4x^2 - 32 x + 55 >0$
$=>4(x - 4)^2 >9$
$=>x > 6.25$
So we seek solution for $\lfloor x \rfloor$ = 6 or 7 or 8
for $6 , 4x^2 = 240-51 = 189$ A solution
for $7, 4x^2 = 229$ A solution
for $8, 4x^2 = 269$ A solution
so solutions are $\frac{\sqrt{189}}{2},\frac{\sqrt{229}}{2}, \frac{\sqrt{269}}{2}$

 

[anemone]

Hi kaliprasad,

Spoiler

Nice try but the solutions aren't complete...:(

 

[kaliprasad]

Hi kaliprasad,

Spoiler

Nice try but the solutions aren't complete...:(

you are right. my mistake

Spoiler

Forgot the solution to the left

$4(x-4)^2 > 9$ also gives $x < 2.5$
$2(x-5)^2 <49$ give $x > 1.5$
so take $\lfloor x\rfloor =2$ to get $4x^2 = 29$ which gives another solution $\frac{\sqrt{29}}{2}$

 

[Greg]

Find all real solutions for the system \(\displaystyle 4x^2-40\left\lfloor{x}\right\rfloor+51=0\).

Spoiler

As noted by kaliprasad, $x$ is positive.

If $m=\lfloor x\rfloor$ then $x=m+c$ for some $0\le c<1$, so we have

\(\displaystyle 4(m+c)^2-40m+51=0\)

\(\displaystyle 4m^2+8mc+4c^2-40m+51=0\implies4m^2-40m+51=-8mc-4c^2\)

As $m$ and $c$ are both positive, all possible $m$ are integers between the zeros of $4m^2-40m+51=0$.

This gives $m=2,3,4,5,6,7,8$ to check, so we have

$4x^2=29$
$4x^2=69$
$4x^2=109$
$4x^2=149$
$4x^2=189$
$4x^2=229$
$4x^2=269$

Solving the above equations for $x$ and substituting into the original equation we find solutions at

\(\displaystyle x\in\left\{\dfrac{\sqrt{29}}{2},\dfrac{\sqrt{189}}{2},\dfrac{\sqrt{229}}{2},\dfrac{\sqrt{269}}{2}\right\}\)

 

[anemone]

Good job to both of you! And thanks for participating!