F.4 A maths binomial

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! ^^

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