Ecuatie de gradul II
Ecuatie de gradul II
Se da ecuatia \( x^2-5x-2=0 \) cu radacinile \( x_1,x_2 \) . Sa se afle valoarea expresiei : \( E=\frac{2x_1^3-3x_1^2+1}{x_1^3-5x_1^2}+\frac{2x_2^3-3x_2^2+1}{x_2^3-5x_2^2} \) .
. A snake that slithers on the ground can only dream of flying through the air.
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
O remarca. Pentru o ecuatie polinomiala de grad \( n\ge 2 \) , orice putere naturala \( x_s^{n+p}\ ,\ p\in \mathbb N \) a unei
radacini \( x_s\ ,\ s\in\overline {1,n} \) se poate exprima in functie de puterile sale inferioare lui \( n\ , \) adica \( x_s^{n+p}=\sum_{k=0}^{n-1}A_kx_s^{k}\ . \)
De exemplu, pentru ecuatia de gradul doi \( x^2-mx+n=0 \) avem \( x_k^{2+p}=mx_k^{1+p}-nx_k^p\ ,\ k\in\overline {1,2}\ ,\ p\in\mathbb N \) .
\( x_k^2=mx_k-n \)
\( x_k^3=mx_k^2-nx_k=m\left(mx_k-n\right)-nx_k=\left(m^2-n\right)x_k-mn \)
\( x_k^4=mx_k^3-nx^2_k=m\left[\left(m^2-n\right)x_k-mn\right]-n\left(mx_k-n\right)=m\left(m^2-2n\right)x_k+n\left(n-m^2\right) \)
etc.
radacini \( x_s\ ,\ s\in\overline {1,n} \) se poate exprima in functie de puterile sale inferioare lui \( n\ , \) adica \( x_s^{n+p}=\sum_{k=0}^{n-1}A_kx_s^{k}\ . \)
De exemplu, pentru ecuatia de gradul doi \( x^2-mx+n=0 \) avem \( x_k^{2+p}=mx_k^{1+p}-nx_k^p\ ,\ k\in\overline {1,2}\ ,\ p\in\mathbb N \) .
\( x_k^2=mx_k-n \)
\( x_k^3=mx_k^2-nx_k=m\left(mx_k-n\right)-nx_k=\left(m^2-n\right)x_k-mn \)
\( x_k^4=mx_k^3-nx^2_k=m\left[\left(m^2-n\right)x_k-mn\right]-n\left(mx_k-n\right)=m\left(m^2-2n\right)x_k+n\left(n-m^2\right) \)
etc.
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mihai miculita
- Pitagora
- Posts: 93
- Joined: Mon Nov 12, 2007 7:51 pm
- Location: Oradea, Romania
\( \mbox{Nu strica putina teorie in plus! Aici merge insa mai simplu: }\\
x_1^2-5x_1-2=0\Rightarrow x_1^3-5x_1^2=2x_1 \mbox{ si analog: } x_2^3-5x_2=2x_2\Rightarrow E=\frac{2x_1^3-3x_1^2+1}{x_1^3-5x_1^2}+\frac{2x_2^3-3x_2^2+1}{x_2^3-5x_2^2}=
\frac{2.(x_1^3-5x_1^2)+7.x_1^2+1}{x_1^3-5x_1^2}+\frac{2.(x_2^3-5x_2^2)+7x_2^2+1}{x_2^3-5x_2^2}=\\
=4+\frac{7.x_1^2+1}{x_1^3-5x_1^2}+\frac{7x_2^2+1}{x_2^3-5x_2^2}=4+\frac{7.x_1^2+1}{2x_1}+\frac{7.x_2^2+1}{2x_2}=4+\frac{7}{2}.(x_1+x_2)+\frac{x_1+x_2}{2x_1x_2}. \)
x_1^2-5x_1-2=0\Rightarrow x_1^3-5x_1^2=2x_1 \mbox{ si analog: } x_2^3-5x_2=2x_2\Rightarrow E=\frac{2x_1^3-3x_1^2+1}{x_1^3-5x_1^2}+\frac{2x_2^3-3x_2^2+1}{x_2^3-5x_2^2}=
\frac{2.(x_1^3-5x_1^2)+7.x_1^2+1}{x_1^3-5x_1^2}+\frac{2.(x_2^3-5x_2^2)+7x_2^2+1}{x_2^3-5x_2^2}=\\
=4+\frac{7.x_1^2+1}{x_1^3-5x_1^2}+\frac{7x_2^2+1}{x_2^3-5x_2^2}=4+\frac{7.x_1^2+1}{2x_1}+\frac{7.x_2^2+1}{2x_2}=4+\frac{7}{2}.(x_1+x_2)+\frac{x_1+x_2}{2x_1x_2}. \)
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Evident, esti neinspirat sa nu speculezi particularitatea problemei (ca in cazul de fata)
ca sa nu zic nebun si sa mergi in pas cadentat pe drumul indicat de vreo metoda generala ...
Oi fi tu elev bine instruit de vreun tambur major, dar esti lipsit de initiativa in situatii particulare.
Asa ca este bine zis - "in general nu strica putina teorie in plus, insa aici merge mai simplu" ....
Noapte buna si numai bine, dl. profesor Miculita.
ca sa nu zic nebun si sa mergi in pas cadentat pe drumul indicat de vreo metoda generala ...
Oi fi tu elev bine instruit de vreun tambur major, dar esti lipsit de initiativa in situatii particulare.
Asa ca este bine zis - "in general nu strica putina teorie in plus, insa aici merge mai simplu" ....
Noapte buna si numai bine, dl. profesor Miculita.