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

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.

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.

## Comments

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

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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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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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. ]

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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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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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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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