Find $\lfloor k^4\rfloor$ where $k^3 - 5k^2 + 2 = 0$.

Question

Problem :

Find a value of $\lfloor k^4 \rfloor$ where $k$ is the biggest real root of equation $k^3 - 5k^2 + 2 = 0$.

Can't use calculator.

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Let $f(x) = x^3 - 5x^2 + 2$. Then $f(4)f(5)<0$. so $k\in(4,5)$.

And I bounded more like : $$\frac{9}{2}<k<5.$$

But this is not enough to Find $\lfloor k^4 \rfloor$ because I have $$410<\left(\frac{9}{2}\right)^4 < k^4<5^4=625$$ and this is too rough to find $\lfloor k^4 \rfloor$.

I tried Newton's method with $x_0 = \frac{9}{2}$ but this is so messy, I got $x_1=\frac{612}{126}.$

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So, I think I need some tight inequality for $k$ or some tricky technique, But I don't know how should I do. Thanks for any help.

Answer 1

If $a<b<c$ are roots of the equation, so by the Viete's theorem $a^4+b^4+c^4=585$ and since for another roots $x_i$ we have $|x_i|<0.7,$ we obtain the answer: $[c^4]=584.$

Indeed, $a+b+c=5,$ $ab+ac+bc=0$ and $abc=-2$.

Thus, $$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)=$$ $$=(a+b+c)^4+4(a+b+c)abc=625-40=585.$$

Can you end it now?