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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: $$x = re^{i\theta}$$ ...
Bertrand87's user avatar
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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?&...
Andrew D. Hwang's user avatar
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 ...
corindo's user avatar
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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 ...
Pynex's user avatar
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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 ...
mengdie1982's user avatar
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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 ...
ράτ's user avatar
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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 ...
Brendan Mitchell's user avatar
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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 ...
rschwieb's user avatar
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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 : ...
Jean Marie's user avatar
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Pell numbers $a_n$, prove that $\sqrt{2a_{2n-1}\pm\sqrt{2}}\in\Bbb Q[e^{i\pi/8}]$

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}-(...
hbghlyj's user avatar
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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 ...
Gwen's user avatar
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2 votes
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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} \...
Thành Nguyễn's user avatar

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