I have asked this before,but nobody could s olve it !,by use of identities and equations,:a^5 +b^5 =64 ,a+b=4,

Question

find:a ,b =?

Answer 1

a^5 +b^5 =64 ....(1)
a+b=4 ................(2)
so from (2) b = 4 - a
put this in (1)
a^5 +(4-a)^5 =64
expanding this reduces it to a 4th degree polynomial
I'll leave out the horrible working and give the result
a^4 - 8a^3 + 32a^2 - 64a +48 = 0.......(3)
now you could go to the quartic formula for a solution but this would be awful
as everyone else has guessed you know a = 2 is a solution
so (a-2) is a factor, polynomial division yields
a^3 - 6a^2 + 20a - 24
note a = 2 solves this too
so polynomial division by (a-2) again yields
a^2 - 4a + 12
and by the quadratic formula
a^2 - 4a + 12 = (a - (2 + i√8))(a - (2 - i√8))
so all up equation (3)
a^4 - 8a^3 + 32a^2 - 64a +48 = 0
factors to
(a - (2 + i√8)) (a - (2 - i√8)) (a-2)^2 = 0


so the only possible solutions are

a=2, b=2 ......... [[ this is the only real solution ]]

a = 2 + i√8 , b = 2 - i√8

a = 2 - i√8 , b = 2 + i√8

where I used (2) to calculate b in each case


the end
.

Answer 2

a=b=2

Answer 3

There are 3 solutions for (a,b) ; two of the solutions are complex numbers.

Here are the steps to solve for one of the variables (I solved for 'b'):

•a=4-b

•a^5 + b^5 = 64

•(4-b)^5 + b^5 = 64

•b^5 – (b-4)^5 = 64

•b^5 – (b^5 -5*4b^4 + 10*16b^3 -10*64b^2 +5*256b – 1024) = 64

•20b^4 - 160b^3 +640b^2 -1280b + 1024 = 64

•20b^4 - 160b^3 +640b^2 -1280b + 960 = 0

•b^4 - 8b^3 + 32b^2 -64b + 48= 0

•(b-2)(b^3 -6b^2 + 20b -24) =0

•(b-2)(b-2)(b^2 -4b +12) =0

Hence, b =2 or b = 2+ (i)(squareroot8), or b = 2 - (i)(squareroot8).

The three solutions are thus:
(2,2),
(2+ (i)(squareroot8), 2- (i)(squareroot8)),
and
( 2+ (i)(squareroot8), 2- (i)(squareroot8)).

* * * NOTE after posting: I just reviewed the other answers and see that the previous answer is the same approach.

Answer 4

A and b both equal 2. There are only so many whole numbers that you can add together to = 4, 0 and 4 or 1 and 3 or 2 and 2. Obviously if you're working with decimals or fractions it gets harder, but o well. Anyways, 4^5 + 0^5 = 1024, not the answer we want. 1^5 + 3^5 = 244. Not it either. Which leaves us with 2^5 + 2^5, which = 64! yay! :)

Answer 5

a=2 b=2.

Answer 6

You should have a cool mind to understand this. this is simple algebra.

Use pascal triangle and find the equation for (a+b)^5

(a+b)^5= a^5 +5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5
=a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)
in the above equation the value of a^3+b^3 is to be found out

Use pascal triangle and find the equation for (a+b)^3
(a+b)^3 = a^3+3a^2b+3ab^2+b^3
= a^3+b^3+3ab(a+b)
a^3+b^3 = (a+b)^3-3ab(a+b)
now sustitute the value of a+b = 4
a^3+b^3 = (4)^3-3ab(4)
=64 - 12ab

now sustitute the value of a+b = 4 & a^3+b^3 =64 - 12ab in the equation (a+b)^5
4^5=64+5ab(64-12ab)+10a^2b^2(4)
1024=64 + 320ab -60a^2b^2+40a^2b^2
20a^2b^2-320ab-64+1024=0......simplify this
a^2b^2-16ab+48 = 0
(ab-12)(ab-4)=0
ab=12 or ab=4

now you know (a-b)^2=a^2+b^2-2ab
...now add & subtract 2ab
=a^2+b^2-2ab-2ab+2ab
=(a+b)^2-4ab
(a-b)^2=(a+b)^2-4ab
inthis equation substitute the value of a+b & ab
case: 1 when ab=12
(a-b)^2=4^2-4.12=-48
so value of a-b can not be found, because sqrt(-48) is imaginary

case: 2 when ab=4
(a-b)^2=4^2-4.4=0
a-b = 0
a=b
Since a+b = 4 & a=b, value of a=b=2

That is your answer

Answer 7

these guys are right. 2^5 = 32. so 32 + 32 = 64; and 2+2 = 4

Answer 8

a^5 +b^5 =64 , a⁵ + b⁵ = 2(2)⁵
a+b=4

a = 4 - b

(4 - b)⁵ + b⁵ = 2(2)⁵ etc

this is too tedious to solve by conventional methods.

the degree of the equation is 5th so there are at least 5 solutions and I can think of 7 of the top of my head.

a = 2, b = 2
2⁵ + 2⁵ = 2(2)⁵

a = 0 , b = 2 x fifth root of 2, and vice versa for a & b.
a = 1, b = fifth root of 63, and vice versa for a & b.
a = -1, b = fifth root of 65, and vice versa for a & b.

Answer 9

a^5 = 32, b^5 = 32
a^4 = 16, b^4 = 16
a^3 = 8, b^3 = 8
a^2 = 4, b^2 = 4
a = 2, b = 2

Answer 10

2^5 = 32

a + b = 4
a^5 + b^5 = 64
a = 2 and b = 2.

a^5 + (4-a)^5 = 64
a^5 + (-a^5 + 20a^4 -160a^3 + 640a^2 -1280a +1024) = 64
20a^4 -160a^3 + 640a^2 -1280a +1024 = 64

you can solve this by looking at the information at this link:
http://books.google.com/books?id=d4a99Fls3CcC&pg=PA656&lpg=PA656&dq=roots+of+a+quartic+funtion&source=web&ots=DPYxA9I_WK&sig=R59nZFmlMW0HH8g8waMcKDZ5I4M