Rozwiąż równanie \(\displaystyle{ (1+x+x^{2})^{2}= \frac{a+1}{a-1}(1+x^{2} + x^{4})}\) z niewiadomą \(\displaystyle{ x}\), jeśli \(\displaystyle{ \left|a \right| \ge 2}\)
Dzięki
Równanie wielomianowe
-
bzyk12
- Użytkownik
- Posty: 327
- Rejestracja: 18 lut 2009, o 12:19
- Płeć: Mężczyzna
- Lokalizacja: Oświęcim/Wawa
- Podziękował: 39 razy
- Pomógł: 43 razy
Równanie wielomianowe
\(\displaystyle{ (a-1)(1+x+x ^{2}) ^{2} =(a+1)[(1+x+x ^{2}) ^{2}-2x( 1+x+x ^{2})]}\)
\(\displaystyle{ (a-1)(1+x+x ^{2}) ^{2} =(a+1)(1+x+x ^{2}) (1-x+x ^{2})}\)
\(\displaystyle{ (1+x+x ^{2})[(a-1)(1+x+x ^{2})-(a+1)(1-x+x ^{2})] =0}\)
\(\displaystyle{ (1+x+x ^{2})(a+ax+ax ^{2}-1-x-x ^{2}-a+ax-ax ^{2}-1+x-x ^{2} )=0}\)
\(\displaystyle{ (1+x+x ^{2})(x ^{2}-ax+1)=0}\)
\(\displaystyle{ \Delta=a ^{2}-4}\)
\(\displaystyle{ x _{(a)1}= \frac{a+ \sqrt{a ^{2}-4 } }{2}}\)
\(\displaystyle{ x _{(a)2}= \frac{a- \sqrt{a ^{2}-4 } }{2}}\)
\(\displaystyle{ (a-1)(1+x+x ^{2}) ^{2} =(a+1)(1+x+x ^{2}) (1-x+x ^{2})}\)
\(\displaystyle{ (1+x+x ^{2})[(a-1)(1+x+x ^{2})-(a+1)(1-x+x ^{2})] =0}\)
\(\displaystyle{ (1+x+x ^{2})(a+ax+ax ^{2}-1-x-x ^{2}-a+ax-ax ^{2}-1+x-x ^{2} )=0}\)
\(\displaystyle{ (1+x+x ^{2})(x ^{2}-ax+1)=0}\)
\(\displaystyle{ \Delta=a ^{2}-4}\)
\(\displaystyle{ x _{(a)1}= \frac{a+ \sqrt{a ^{2}-4 } }{2}}\)
\(\displaystyle{ x _{(a)2}= \frac{a- \sqrt{a ^{2}-4 } }{2}}\)