I know that I can easly solve this with the $4$th degree's equation, but isn't there a smarter way? It is an olympiad's problem so it shouldn't be a formula, but more find a formula… I have remarked that it is equal to $(x^2+1)\cdot(x+1)\cdot(x-1)+4x=0$ It isn't just solving the equation, I want to know if it can be solved otherwise than that… Could someone explain why this is equal to $(x^2+1)^2-(\sqrt{2}(x-1))^2$ please ?
[Math] How to solve this equation; $x^4+4x-1=0$
algebra-precalculusquartics
Related Solutions
Since $\frac{15x-7}{5}$ is required to be an integer, it is of the form $\frac{2+5j}{15}$ for $j$ an arbitrary integer. Thus, the equation to solve becomes $$ \left\lfloor\frac{6\big(\frac{2+5j}{15}\big)+5}{8}\right\rfloor = \frac{15\big(\frac{2+5j}{15}\big)-7}{5}\qquad\text{which simplifies to}\quad \left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor = j-1 $$ Now observe that the function on the right hand side increases quicker than the function on the left hand side. So we can solve the question by finding a maximal solution to the inequality $$ j-1\leq\left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor $$
A warning: But here we should be careful, because a maximal solution to this inequality is not necessarily a solution to our equality. Also, there might be more solutions than just the maximal solution $j_0$ to this inequality. But since the left hand side and the right hand side are linear in $j$, the set of solutions forms an interval in $\mathbb{Z}$. This means that to give all the solutions, it suffices to find the largest integer below $j_0$ for which the inequality is strict.
Note that for any $n\in\mathbb{Z}$ and $a\in\mathbb{R}$, the inequality $n\leq\lfloor a\rfloor$ holds if and only if $n\leq a$ holds. This helps, since now we can continue to compute $$ \begin{split} j-1 \leq \left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor & \Leftrightarrow j-1 \leq \frac{\frac{4}{5}+2j+5}{8}\\ & \Leftrightarrow 8j-8\leq \frac{4}{5}+2j+5\\ & \Leftrightarrow 10j\leq 23\\ & \Leftrightarrow j\leq 2. \end{split} $$ Therefore, we get the maximal solution for the equation by substituting $j=2$ in $x=\frac{2+5j}{15}$, i.e. $x=\frac{4}{5}$. This is indeed a solution to the equality. As we have noted before, $\frac{4}{5}$ is not necessarily the only solution. Indeed, it is easy to check that $j=1$ also gives a solution to the equation. But for $j\leq 0$ the inequality is strict, so the only solutions to the equation are $x=\frac{4}{5}$ and $x=\frac{7}{15}$.
Here is a link to the answer to another question, where the OP had to deal with the floor function. Maybe you'll find it useful.
For the first one, if the product of three factors is 0 then at least one of these is 0; so you have to solve three (easy) equations.
For the second one, write it as $x(x^3−x^2−x+1)=0$ and note that the sum in parentheses is 0 when $x=1$, so you may factor $x^3−x^2−x+1 = (x-1)(x^2+Ax+1)$. Find the value of $A$ and solve the 2nd degree equation
Best Answer
You may write it as $$ x^4+4x-1=(x^2+1)^2-2(x-1)^2 = (x^2+1)^2-(\sqrt{2}(x-1))^2$$ and factorize.