Is it possible to turn $\sqrt\frac12$ into $\sqrt2$? $ \frac{3}{4}\sqrt{\frac{1}{2} } = \frac{3}{8}\sqrt{{2} }$: What is between the left/right sides?

score 82 · Answer 11 · 2023-11-26 21:08:25

82

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Tedious rational integral

Published on 26 Nov 2023 - 21:08

score 300 · Answer 12 · 2023-11-28 08:36:31

300

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How to prove $\sqrt{2a+3bc}+\sqrt{2b+3ca}+\sqrt{2c+3ab}\ge 3\sqrt{5}$?

Published on 28 Nov 2023 - 8:36

score 47 · Answer 13 · 2023-11-30 00:37:38

47

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Confused about the extraneous root of $\frac{x \sqrt{A^2 - x^2} + x}{x^2 - \sqrt{A^2 - x^2}}$

Published on 30 Nov 2023 - 0:37

score 54 · Answer 14 · 2023-12-02 15:29:59

54

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What are the solutions for the inequality $x^2 \geq 60$

Published on 02 Dec 2023 - 15:29

score 82 · Answer 15 · 2023-12-04 15:11:05

82

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If $\alpha^3-\alpha+1=0$ then $\sqrt{3}\neq a\alpha^2+b\alpha+c$ for $a$, $b$, $c\in \mathbf{Q}$

Published on 04 Dec 2023 - 15:11

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Is it possible to turn $\sqrt\frac12$ into $\sqrt2$? $ \frac{3}{4}\sqrt{\frac{1}{2} } = \frac{3}{8}\sqrt{{2} }$: What is between the left/right sides?

Tedious rational integral

How to prove $\sqrt{2a+3bc}+\sqrt{2b+3ca}+\sqrt{2c+3ab}\ge 3\sqrt{5}$?

Confused about the extraneous root of $\frac{x \sqrt{A^2 - x^2} + x}{x^2 - \sqrt{A^2 - x^2}}$

What are the solutions for the inequality $x^2 \geq 60$

If $\alpha^3-\alpha+1=0$ then $\sqrt{3}\neq a\alpha^2+b\alpha+c$ for $a$, $b$, $c\in \mathbf{Q}$

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