F.4 A maths binomial @ 新聞雲

標題:

F.4 A maths binomial

發問:

given that (a+b)^4 - 4ab(a+b)^2 +2a^2*b^2=a^4+b^4. and ais a positive number and b is a negative number satisfying a + b =1 and a^4+b^4 =97 . using the result, show that ab = -6. Hence , find the values of a and b.

最佳解答:

since a+b = 1 and a^4+b^4 = 97, from the given, (1)^4 - 4ab(1)2 +2a2*b2 = 97 1 - 4ab + 2(ab)2 = 97 2(ab)2 - 4(ab) - 96 = 0 (ab)2 - 2(ab) - 48 = 0 (ab-8)(ab+6) = 0 ab = 8 (Rejected since a＞0 and b＜0 and so ab＜0) or ab = -6 So, ab = -6 a = -6/b Put into a + b = 1, -6/b + b = 1 -6 + b2 = b b2 - b - 6 = 0 (b-3)(b+2) = 0 b = 3 (Rejected since b＜0) or b = -2 So b = -2 a = -6/(-2) = 3 Hope it helps! ^^

其他解答:

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a + b = 1----------(1) a^4 + b^4 = 97------------(2) (a+b)^4 - 4ab(a+b)2 +2a2b2 = a^4+b^4 (a+b)^4 - 4ab(a+b)2 +2(ab)2 = 97 (1)^4 - 4ab(1)2 + 2(ab)2 = 97 1 - 4ab + 2(ab)2 = 97 2(ab)2 - 4ab - 96 = 0 (ab)2 - 2ab - 48 = 0 (ab-8)(ab+6) = 0 ab = 8 (rejected) or ab = -6 ∴ ab = -6 a + b = 1----------(1) a^4 + b^4 = 97------------(2) ab = -6---------------(3) From(1),a = 1 - b Subs a = 1 - b into (3) (1-b)b = -6 b - b2 + 6 = 0 b2 - b - 6 = 0 (b-3)(b+2)=0 b = 3 (rejected) or b = -2 When b = -2 , a = 3|||||(a+b)^4 - 4ab(a+b)^2 +2a^2*b^2=a^4+b^4 1 -4ab*1^2+2(ab)^2=97 2(ab)^2-4ab-96=0 x^2-2x-48=0 such that x=ab x =-6 or 8(rej) as a is a positive number and b is a negative number a=3,b=-2 as ab=-6 ,a+b =1|||||Put a^4+b^4 =97 and a + b =1 into (a+b)^4 - 4ab(a+b)^2 +2a^2*b^2=a^4+b^4 1-4ab+2(ab)^2=97 (ab)^2 -2ab -48=0 (ab-8)(ab+6)=0 ab=-6 (rej, ab=8 because a>0 and b<0) put b=-6/a into a+b=1 a-6/a=1 a^2-a-6=0 (a-3)(a+2)=0 a=3 (rej. a=-2 as a>0) Hence b=-6/3=-2