(a) Expand (2m+n)^4 + (2m-n)^4.
(b) Hence, evaluate (2+√3)^4 + (2-√3)^4.
(c) Form a quadratic equation in x whose roots are (2+√3)^4 and (2-√3)^4.
Binomial Expansion(5)
2010-03-05 3:05 am
回答 (1)
2010-03-05 3:20 am
✔ 最佳答案
(a) (2m+n)^4 + (2m-n)^4=4C0(2m)^4+4C1(2m)^3(n)+4C2(2m)^2(n)^2+4C3(2m)(n)^3+4C4(n)^4 + 4C0(2m)^4+4C1(2m)^3(-n)+4C2(2m)^2(-n)^2+4C3(2m)(-n)^3+4C4(-n)^4
=2(16m^4+24m^2n^2+n^4)
=32m^4+48m^2n^2+2n^4
(b) Let m=1 and n=√3
32(1)^4+48(1)^2( √3)^2+2(√3)^4
=32+144+18
=194
(c) Let the equation be ax^2 + bx + c=0
Sum of roots= -b/a =[ (2+√3)^4 + (2-√3)^4] = 194
Product of roots=c/a= (2+√3)^4x (2-√3)^4=(4-3)^4=1
The equation= x^2-194x+1=0
收錄日期: 2021-04-23 18:24:08
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https://hk.answers.yahoo.com/question/index?qid=20100304000051KK01060