Sa se rezolve ecuatiile peste \( \mathbb R \) :
\( 1\ \odot\ \ x^4+4x=1 \) .
\( 2\ \odot\ \ x^4+4x^3+6x^2+8x+4=0 \) .
Cateva ecuatii polinomiale.
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Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
\( 1\ \odot\ x^4+4x=1\ \Longleftrightarrow\ x^4+2x^2+1-2x^2+4x-2=0\ \Longleftrightarrow\ (x^2+1)^2-2(x^2-2x+1)=0\ \Longleftrightarrow\ (x^2+1)^2-\left\( (x-1)\sqrt 2\right\)^2=0 \)
\( \Longleftrightarrow\ (x^2+x\sqrt 2+1-\sqrt 2)(x^2-x\sqrt 2+1+\sqrt 2)=0 \) cu solutiile reale \( \overline{\underline{\left\|\ x_{1,2}=-\frac{1}{\sqrt 2}\ \pm\ \sqrt{\frac 12(2\sqrt 2-1)}\ \right\| \) .
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\( 2\ \odot\ x^4+4x^3+6x^2+8x+4=0\ \Longleftrightarrow\ x^4+4x^2+4+4x^3+4x^2+8x-2x^2=0\ \Longleftrightarrow\ (x^2+2x+2)^2-2x^2=0 \)
\( \Longleftrightarrow\ (x^2+x(2-\sqrt 2)+2)(x^2+x(2+\sqrt 2)+2)=0 \) cu solutiile reale \( \overline{\underline{\left\|\ x_{1,2}=-1-\frac{1}{\sqrt 2}\ \pm\ \sqrt{\frac12(2\sqrt 2-1)}\ \right\| \) .
\( \Longleftrightarrow\ (x^2+x\sqrt 2+1-\sqrt 2)(x^2-x\sqrt 2+1+\sqrt 2)=0 \) cu solutiile reale \( \overline{\underline{\left\|\ x_{1,2}=-\frac{1}{\sqrt 2}\ \pm\ \sqrt{\frac 12(2\sqrt 2-1)}\ \right\| \) .
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\( 2\ \odot\ x^4+4x^3+6x^2+8x+4=0\ \Longleftrightarrow\ x^4+4x^2+4+4x^3+4x^2+8x-2x^2=0\ \Longleftrightarrow\ (x^2+2x+2)^2-2x^2=0 \)
\( \Longleftrightarrow\ (x^2+x(2-\sqrt 2)+2)(x^2+x(2+\sqrt 2)+2)=0 \) cu solutiile reale \( \overline{\underline{\left\|\ x_{1,2}=-1-\frac{1}{\sqrt 2}\ \pm\ \sqrt{\frac12(2\sqrt 2-1)}\ \right\| \) .