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1 vote
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Do equations $\sqrt{x}=-2$ and $\sqrt[3]{x}=-2$ have complex solutions?

I think it is a good idea to solve your first equation $\sqrt{x}=-2$ from first principles: We know that any complex number different from zero has a polar representation: $$x = re^{i\theta}$$ ...
• 3,257
1 vote

Do equations $\sqrt{x}=-2$ and $\sqrt[3]{x}=-2$ have complex solutions?

$\newcommand{\Sqrt}{\sqrt{\rule{0pt}{4pt}\quad}}$tl; dr 1: Reasonable as it sounds, the question as currently worded ("Do equations $\sqrt{x} = -2$ and $\sqrt[3]{x} = -2$ have complex solutions?&...
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1 vote

Do equations $\sqrt{x}=-2$ and $\sqrt[3]{x}=-2$ have complex solutions?

$\sqrt{x} = -2$ doesn't have a solution in $\mathbb{R}$ because when we define the real-valued square-root function $\sqrt{\cdot}$ to be the inverse of the square function $x\mapsto x^2$, we choose to ...
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0 votes

Multiple definitions of casus irreducibilis

Like a lot of terms in mathematics, the meaning of casus irreducibilis falls in the category of "it depends". The historical importance of casus irreducibilis was that Cardano's method ...
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Prove that $\forall x>0, \frac {x-1}{\ln(x)} \geq \sqrt{x}$.

We may assume $x\ge 1$. If $x<1$, then put $\frac{1}{x}$ into the inequality, which keep the same form. Since $f(t):=\frac{1}{t}$ is convex on $(0,+\infty)$, by Hermite-Hadamard inequality, it ...
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3 votes

An infinite nested radical

Let $$x={\sqrt {4+\sqrt {4+\sqrt {4-\sqrt {4+\sqrt {4+\sqrt {4- ......}}}}}}}$$ then $x={\sqrt {4+\sqrt {4+\sqrt {4-x}}}}$ Now we just need to solve for $x$ Start by eliminating the square root of ...
• 899
1 vote

Sign of square root in $\sqrt{\frac{4}{9}}$

Previous answers and comments are correct, but to elaborate/rephrase a little further: The reason why the square root of 4/9 (or any other positive number) is positive and not negative is because we ...
2 votes
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Is there a non-commutative ring with unity $R$ such that its Jacobson radical is not a two-sided maximal ideal?

The Jacobson radical is not often maximal, because $R/J(R)$ does not have to be a simple ring. It would seem you have answered your own question already, because your example is fine. Given any two ...
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Other methods of solving this question: finding $2y^4-8y^3-5y^2+26y-28$ for $y=1+\sqrt2+\sqrt3$

I would like here to stress the help that a software like SAGE can bring, using the concept of minimal polynomial. Here is the program : ...
• 83.1k
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Substituting $$2a_{2n-1}=\frac{1}{\sqrt{2}}\left((1+\sqrt{2})^{2 n-1}-(1-\sqrt{2})^{2 n-1}\right)$$ into left side: $$\sqrt{2a_{2n-1}+(-1)^n\sqrt{2}}=\sqrt{\frac{1}{\sqrt{2}}\left((1+\sqrt{2})^{2n-1}-(... • 2,832 1 vote Other methods of solving this question: finding 2y^4-8y^3-5y^2+26y-28 for y=1+\sqrt2+\sqrt3$$\begin{align*} y-1-\sqrt 2=\sqrt 3 \\ \implies y^2+1+2-2y-2\sqrt 2y+2\sqrt 2=3\\ \implies y^2-2y=2\sqrt2(y-1)=2\sqrt 2(\sqrt 2+\sqrt 3) \\ \implies y^2-2y=4+2\sqrt 6\end{align*}$$Input this value ... • 2,668 2 votes Accepted Other methods of solving this question: finding 2y^4-8y^3-5y^2+26y-28 for y=1+\sqrt2+\sqrt3$$ y = 1 + \sqrt{2} + \sqrt{3} \Longrightarrow (y - 1)^2 = (\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{6} $$Bring 5 to the LHS, and then square both side again,$$ (y - 1)^2 - 5 = 2\sqrt{6} \...
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