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Actually, the solution is simpler than you might think.

Rearranging the equation:
$40\lfloor x\rfloor=4x^2+51$ $40(x-1)<4x^2+51\leq40x$

From the lower limit,
$4x^2+51>40(x-1)$$(x-\frac{7}{2})(x-\frac{13}{2})>0$

($x<\frac{7}{2}$ and $x< \frac{13}{2}$)
OR
($x>\frac{7}{2}$ and $x>\frac{13}{2}$ )
which imply $x<\frac{7}{2}$ OR $x>\frac{13}{2}$

Similarly, from upper limit,
$4x^2+51\leq 40x$$(x-\frac{17}{2})(x-\frac{3}{2})\leq 0$

($x\geq \frac{17}{2}$ and $x \leq \frac{3}{2}$)
OR
($x\leq \frac{17}{2}$ and $x\geq \frac{3}{2}$ )
which imply (an impossible case) OR $\frac{3}{2} \leq x \leq \frac{17}{2}$

Hence, solutions are in the region
$(1.5\leq x<3.5)U( 6.5<x\leq8.5)$

Rearranging the given equation, we get
$x=\frac{\sqrt{40\lfloor x \rfloor-51}}{2}$

Solving in each region,

$@ 1.5\leq x<2, \lfloor x \rfloor=1$
x has no real solutions

$@ 2\leq x<3, \lfloor x \rfloor=2$$x=\frac{\sqrt{29}}{2}$

$@ 3\leq x<3.5, \lfloor x \rfloor=3$$x=\frac{\sqrt{69}}{2}$
Which is a contradiction (as it is not in the region)

(Note: I may even have done some stupid mistake, please verify)

$@ 6.5<x<7, \lfloor x \rfloor=6$$x=\frac{3}{2}\sqrt{21}$

$@ 7\leq x<8, \lfloor x \rfloor=7$$x=\frac{\sqrt{229}}{2}$

$@ 8\leq x<8.5, \lfloor x \rfloor=8$$x=\frac{\sqrt{269}}{2}$

the solutions are$x=\frac{\sqrt{29}}{2},\frac{3}{2}\sqrt{21},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}$

[Please comment, if you find any of the assumptions to be wrong or questionable. ]

@Harish Sasikumar
–
I used the Shreedhacharya rule. it gave me two roots, one is 3/2 as mentioned in your answer. the other was 17/2. I am an amateur at math and might have made some mistake. Joined brilliant.org recently to improve my problem solving skill.

@Silver Vice
–
OK. Understood the mistake. You took it as a quadratic equation and solved. That's how you got those solutions (3/2 and 17/2).

However it is NOT a quadratic equation. There is a 'floor' function in between which is represented as $\lfloor x \rfloor$ and is defined as the the integral part of a real number. For example
$\lfloor 3.4 \rfloor =3$

Keep doing problems Brilliant. It's an awesome source of problems. Best of luck :).

obviously [x]≥2 as $2x=\sqrt{40[x]-51}$ and x real.
put x =[x]+y where y is frac part ≥0 and <1.so
$4([x]+y)^2-40[x]+51=4y^2+8[x]y+(4[x]^2-40[x]+51)=0$
due to y range:(the other one will always <0 on the restrictions)
$0≤\dfrac{-2[x]+\sqrt{40[x]-51}}{2}<1$$0≤-2[x]+\sqrt{40[x]-51}<2$$2[x]≤\sqrt{40[x]-51}<2+2[x]$$4[x]^2≤40[x]-51<4[x]^2+8[x]+4$$4[x]^2-40[x]+51≤0<4[x]^2-32[x]+55$
different cases we get(since all int≥2):
$2≤[x]≤8 \cap [x]≤2(\text{squeezes with [x]≥2 to get [x]=2}) \quad or \quad [x]≥6$
so $\begin{cases}
[x]=2\\
6≤[x]≤8\end{cases}$
just check all 4 cases to get respective y and x.

All the real roots of the given equation will lie in (1,9). This can be easily figured out once you draw the graphs: $y=4x^2+51$ and $y=40\lfloor x \rfloor$ [because the points at which these two graphs will intersect will be the solutions of the given equation] and observe that the later one will never be able to catch the former one.

So now let us look at our equation, which can be rewritten as $4x^2+51=40 \lfloor x \rfloor$. Looking at the RHS we conclude that it is always an integer. Therefore for the equation to be correct LHS must be an integer too.

Now put $\lfloor x \rfloor = i$ in the RHS, where $i$ is an integer lying in $(1,9)$ and find the value of $x$. To be more explicit, only those values of $x$ will be considered as the roots of the given equation which will satisfy $\lfloor x \rfloor = i$ simultaneously.

for ease write the equation as$(2x)^2=40[x]-51$ by definition the floor function is an integer, implying 40[x]-51 is also an integer which also implies $(2x)^2$ is an integer

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestActually, the solution is simpler than you might think.

Rearranging the equation: $40\lfloor x\rfloor=4x^2+51$

$40(x-1)<4x^2+51\leq40x$

From the lower limit, $4x^2+51>40(x-1)$ $(x-\frac{7}{2})(x-\frac{13}{2})>0$

($x<\frac{7}{2}$ and $x< \frac{13}{2}$) OR ($x>\frac{7}{2}$ and $x>\frac{13}{2}$ ) which imply $x<\frac{7}{2}$ OR $x>\frac{13}{2}$

Similarly, from upper limit, $4x^2+51\leq 40x$ $(x-\frac{17}{2})(x-\frac{3}{2})\leq 0$

($x\geq \frac{17}{2}$ and $x \leq \frac{3}{2}$) OR ($x\leq \frac{17}{2}$ and $x\geq \frac{3}{2}$ ) which imply (an impossible case) OR $\frac{3}{2} \leq x \leq \frac{17}{2}$

Hence, solutions are in the region $(1.5\leq x<3.5)U( 6.5<x\leq8.5)$

Rearranging the given equation, we get $x=\frac{\sqrt{40\lfloor x \rfloor-51}}{2}$

Solving in each region,

$@ 1.5\leq x<2, \lfloor x \rfloor=1$ x has no real solutions

$@ 2\leq x<3, \lfloor x \rfloor=2$ $x=\frac{\sqrt{29}}{2}$

$@ 3\leq x<3.5, \lfloor x \rfloor=3$ $x=\frac{\sqrt{69}}{2}$ Which is a contradiction (as it is not in the region)

(Note: I may even have done some stupid mistake, please verify)

$@ 6.5<x<7, \lfloor x \rfloor=6$ $x=\frac{3}{2}\sqrt{21}$

$@ 7\leq x<8, \lfloor x \rfloor=7$ $x=\frac{\sqrt{229}}{2}$

$@ 8\leq x<8.5, \lfloor x \rfloor=8$ $x=\frac{\sqrt{269}}{2}$

the solutions are$x=\frac{\sqrt{29}}{2},\frac{3}{2}\sqrt{21},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}$[Please comment, if you find any of the assumptions to be wrong or questionable. ]

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I found 17/2 to be an answer too.

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How did you get that solution? It doesn't seem correct. $17/2=8.5$

$LHS = 4(8.5)^2 -40\lfloor 8.5 \rfloor +51 =289-320+51=20\neq 0$ It doesn't satisfy the equation

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However it is NOT a quadratic equation. There is a 'floor' function in between which is represented as $\lfloor x \rfloor$ and is defined as the the integral part of a real number. For example $\lfloor 3.4 \rfloor =3$

Keep doing problems Brilliant. It's an awesome source of problems. Best of luck :).

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Can someone explain, why there are 2 'rejectable' solutions coming up as we are solving it ?

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8.20060973342836

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Yes, you are correct. I changed it.

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How about 4 $y$ - 40 $\lfloor \sqrt{y}\rfloor$ + 51 = 0 for all rational y?

y = {7.25, 47.25, 57.25, 67.25}

$z$ - 40 $\lfloor \sqrt{\frac{z}{4}}\rfloor$ + 51 = 0 for all positive integer z!

z = {29, 189, 229, 269}

I guess this is the solution of the equation 4 $x^2$ - 40 $\lfloor x \rfloor$ + 51 = 0.

x = {$\sqrt{\frac{29}{4}}, \sqrt{\frac{189}{4}}, \sqrt{\frac{229}{4}}, \sqrt{\frac{269}{4}}$} = {$\frac{\sqrt{29}}{2}, \frac{3 \sqrt{21}}{2}, \frac{\sqrt{229}}{2}, \frac{\sqrt{269}}{2}$}

This means it doesn't follow only 2 but duplicated into 4. Changed from {$\frac{17}{2}, \frac32$}.

$z$ - 40 $\lfloor -\sqrt{\frac{z}{4}}\rfloor$ + 51 = 0 for all positive integer z could be some more but not found with any.

x = {$\frac{\sqrt{29}}{2}, \frac{3 \sqrt{21}}{2}, \frac{\sqrt{229}}{2}, \frac{\sqrt{269}}{2}$}

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obviously [x]≥2 as $2x=\sqrt{40[x]-51}$ and x real. put x =[x]+y where y is frac part ≥0 and <1.so $4([x]+y)^2-40[x]+51=4y^2+8[x]y+(4[x]^2-40[x]+51)=0$ due to y range:(the other one will always <0 on the restrictions) $0≤\dfrac{-2[x]+\sqrt{40[x]-51}}{2}<1$ $0≤-2[x]+\sqrt{40[x]-51}<2$ $2[x]≤\sqrt{40[x]-51}<2+2[x]$ $4[x]^2≤40[x]-51<4[x]^2+8[x]+4$ $4[x]^2-40[x]+51≤0<4[x]^2-32[x]+55$ different cases we get(since all int≥2): $2≤[x]≤8 \cap [x]≤2(\text{squeezes with [x]≥2 to get [x]=2}) \quad or \quad [x]≥6$ so $\begin{cases} [x]=2\\ 6≤[x]≤8\end{cases}$ just check all 4 cases to get respective y and x.

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A simple standard approach will crush it:

All the real roots of the given equation will lie in (1,9). This can be easily figured out once you draw the graphs: $y=4x^2+51$ and $y=40\lfloor x \rfloor$ [because the points at which these two graphs will intersect will be the solutions of the given equation] and observe that the later one will never be able to catch the former one.

So now let us look at our equation, which can be rewritten as $4x^2+51=40 \lfloor x \rfloor$. Looking at the RHS we conclude that it is always an integer. Therefore for the equation to be correct LHS must be an integer too.

Now put $\lfloor x \rfloor = i$ in the RHS, where $i$ is an integer lying in $(1,9)$ and find the value of $x$. To be more explicit, only those values of $x$ will be considered as the roots of the given equation which will satisfy $\lfloor x \rfloor = i$ simultaneously.

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$(2x)^2- 40 \lfloor x \rfloor +51=0$

Two things to be noticed here:

Therefore, the fractional part of $x$ would be 0.5. Now we can write $x=N+0.5$ and solve it quite easily.

Edit:@Dev Sharma @Aareyan Manzoor @Adarsh Kumar @Akshat Sharda This solution is wrong. Because $2x$ need not be an integer. $(2x)^2\in Z\not \Rightarrow 2x\in Z$. I'll have to think again about an alternative method.Log in to reply

One can easily draw the grpah of $\frac{x^2}{10} + \frac{51}{40}$ and $\left \lfloor x \right \rfloor$ and find the intersection.

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why 2x must be integer?

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for ease write the equation as$(2x)^2=40[x]-51$ by definition the floor function is an integer, implying 40[x]-51 is also an integer which also implies $(2x)^2$ is an integer

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Ooops!I didn't see that !Thanx a lot!

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