# Tag Info

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### Why is cubic function one-one?

If $(x_1-x_2)(x_1^2+x_2^2+x_1x_2)=0$, then as you implied, either $x_1=x_2$ and we are done, or else $x_1\ne x_2$. Now it is enough to show that for every $x_1\in\mathbb{R}$ and for every $x_2\ne x_1$,...
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### Why is cubic function one-one?

Yes, your proof is sufficient. If $x_1$ and $x_2$ satisfy $x_1^2+x_2^2+x_1x_2=0$, then you have shown that both: $x_1,x_2$ have opposite signs because $-x_1x_2=x_1^2+x_2^2>0$. $x_1,x_2$ have the ...
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### Used Chinese Remainder Theorem on this question, and I can't figure what went wrong

All your steps except the last one are correct. It is $$19056 \equiv 1731 \pmod{3465}$$ but not $$19056 \equiv 1236 \pmod{3465}$$
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### Show $\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}+\ldots+\sqrt{a_n^2+b_n^2} \geq \sqrt{\left(a_1+a_2+\ldots+a_n\right)^2+\left(b_1+b_2+\cdots+b_n\right)^2}$

Let $\lVert \cdot \rVert$ be Euclidean norm on $\mathbb{R}^2$, and $v_i=(a_i,b_i)$. Then, the inequality can be written as $\sum_{i=1}^n \lVert v_i \rVert \geqq \lVert \sum_{i=1}^n v_i \rVert$ ,...
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### Numbers that are four times their decimal reverse

This is not a direct answer to the question, but a general way to see numbers defined by digit reversal and linear equality (it should work for any Presburger arithmetic but I don't have formalization)...
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### counting possible coefficient of polynomial

Brute force search in this case is not difficult, because as you noted $r_1 + r_2 = m$, which is less than $20$. We only need to identify pairs of prime numbers (not necessarily distinct) whose sum is ...
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### Solve $\dfrac{x-1}{2}\geq \dfrac{2}{3} \geq 1-\dfrac{x}{6}$

You basically want to find an $x\in \mathbb{R}$ such that $3x-3\geq 4$ and $4\geq 6-x$ both hold. I plotted both inequalities for you : See $6-x\leq 4$ iff $x\geq 2$ but then $3x-3\leq4$ so this x won'...

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### Let $S=\{-1,0,\frac{1}{2},\sqrt{2},2 \}$. Solve for $(1)\ \dfrac{1}{x} \leq \dfrac{1}{2} \ \ \ \ (2)\ 10 \geq\dfrac{5}{2x}$.

It is permissible to directly substitute into the test. However, since the focus is on learning about this category of problems, a thorough understanding of fractional equations with unknowns in the ...
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### $2^x + 3^x + 6^x = x^2$, no. Of solutions. (Without differentiation)

Obviously, $\color{blue}{x=-1}$ is a solution to the given equation $2^x+3^x+6^x=x^2$ as $\frac12+\frac13+\frac16=1$. Let's check out whether other real solutions also exist. First, note that the ...
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### Finding the real roots of an octic polynomial (degree eight).

You're right: the polynomial initially resists simple attempts at finding its real roots through elementary methods like RRT (as a side note; you really only needed to try out the negative factors of ...
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### Solve $16x^4=81$

By the Fundamental theorem of algebra, if $p(x)$ is a polynomial with complex (possibly all real) coefficients of degree $n$, then equation $p(x) = 0$ has precisely $n$ solutions, counted with ...
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### Solve $16x^4=81$
In order to avoid the case of only getting the two real solutions, we should keep in mind that any degree 4 polynomial with complex coefficients (i.e., on $\mathbb{C}[x]$) has 4 roots, counting the ...
### Solve $16x^4=81$
Your first approach is just fine, but remember: $$x^4=\frac{81}{16}\implies x=\frac{3}{2}\times\text{the 4^{th} roots of unity }$$ So x=-\frac{3}{2}\lor x=-\frac{3i}{2}\lor x=\frac{3i}{2}\lor x=\...