4-misol. x6-3x3-2=0 tenglama yechilsin.
Yechish: y=x3 deb belgilab y2-3y+2=0 yordamchi tenglama topa-miz, uning ildizlari y1=1, y2=2.
Natijada x3=1 va x3=2 tenglamalarga ega bo`lamiz. Bular (x- -1)(x2+x+1)=0 va tenglamalarga teng kuchlidir. Birinchisidan, x1=1, ni, ikkinchisidan ni hosil qilamiz.
5-misol. 3x4+26x2-9 bikvadrat uchhad ko`paytuvchilarga ajratilsin.
Yechish: 3x4+26x2-9=0 tenglamani yechamiz: va dan x2=-9 dan x3=3i, x4=-3i ni topamiz va ni hosil qilamiz, yoki hosil bo`ladi (kompleks sonlar to`plamida), Haqiqiy sonlar to`plamida esa bo`ladi
Mashqlar
12 Quyidagi ikki hadli tenglamalarni yeching.
1) x3-8=0 5) x4-16=0
2) x3+8=0 6) x4+16=0
3) x5-32=0 7) x4-81=0
4) x5+32=0 8) x4+81=0
126. Quyidagi uch hadli tenglamalarni yeching:
1) x4+5x2-36=0 5) x4+3x2-18=0
2) x4-8x2-9=0 6) x4+4x2-32=0
3) x4-x2-6=0 7) x4+x2-1=0
4) x4+2x2-15=0 8) x4-2x2+4=0.
127. 1) Ildizlari 1-2i, 2-i bo`lgan haqiqiy koeffitsiyenti to`rtinchi darajali tenglamani tuzing.
2) Ildizlari 1, 1-2i, 2-i bo`lgan haqiqiy koeffitsiyenti beshinchi darajali tenglamani tuzing.
Javoblar: 12 2) x1=-2, 4)
k=0,1,2,3,4; 6) 8) .
126. 2) 3; i; 4) 6)
127. 2) x5-7x4+24x3-48x2+55x-25=0.
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