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O ecuatie care nu are radacini reale (Own).
Posted: Sun Sep 30, 2007 1:08 pm
by Virgil Nicula
Fie numerele reale pozitive \( a\ ,\ b \) pentru care \( a^4\le b^3\ . \)
Sa se arate ca nu exista \( x\in \mathbb R \) astfel incat \( x^4+b=ax\ . \)
Remarca. Cu alte cuvinte, in ipoteza mentionata, \( x\in\mathbb{R}\ \Rightarrow\ x^4+b>ax\ . \)
Posted: Sun Sep 30, 2007 9:53 pm
by Filip Chindea
Evident \( b \neq 0 \). Avem \( x^4 + b = (x^4 + \frac{b}{3}) + (\frac{b}{3} + \frac{b}{3}) \) \( \ge 2\sqrt{(x^4 + \frac{b}{3})(\frac{b}{3} + \frac{b}{3})} \) \( \ge 2\sqrt{2\sqrt{x^4 \cdot \frac{b}{3}} \cdot 2\sqrt{\frac{b}{3} \cdot \frac{b}{3}}} = 4\sqrt{\sqrt{x^4 \cdot \frac{b^3}{27}}} \) \( \ge 4\sqrt{\sqrt{x^4 \cdot \frac{a^4}{27}}} \) \( = a|x| \cdot \frac{4}{\sqrt{\sqrt{27}}} > a|x| \cdot \frac{4}{\sqrt{\sqrt{81}}} = \frac{4a|x|}{3} > a|x| \ge ax. \)
Posted: Thu Sep 09, 2010 12:59 am
by Virgil Nicula
Fie numerele reale pozitive \( a\ ,\ b \) pentru care \( a^4\le b^3\ . \)
Sa se arate ca nu exista \( x\in \mathbb R \) astfel incat \( x^4+b=ax\ . \)
Remarca. Cu alte cuvinte, in ipoteza mentionata, \( x\in\mathbb{R}\ \Rightarrow\ x^4+b>ax\ . \)
Se observa usor ca
\( \begin{array}{c}
(\forall ) x\ge \sqrt[3] a & \Longrightarrow & x^4-ax+b=x\left(x^3-a\right)+b>0\ .\\\\\\\\\
(\forall ) x\le \frac ba & \Longrightarrow & x^4-ax+b=x^4+(-ax+b)>0 .\end{array} \)
Insa
\( a^4\le b^3\Longleftrightarrow \sqrt[3] a\le \frac ba\ . \) In concluzie, pentru orice
\( x\in\mathbb R \) avem
\( x^4-ax+b>0 \) .