Question

\[\begin{align}a+b+c+d&=2\\
a^2+b^2+c^2+d^2&=30\\
a^3+b^3+c^3+d^3&=44\\
a^4+b^4+c^4+d^4&=354\end{align}\]

Answer 1

minor prize goes to the first to solve the system,
major prize goes to the best method

Answer 2

clue; none of \(a,b,c,d\) are equal

Answer 3

hmm nice clue LOL

Answer 4

can we do either using substitution or elimination

Answer 5

it is -1, 2, -3 and 4

Answer 6

second equation is all squared and added. so 1 + 2 + 9 + 16 = 30

All numbers are different you said

So put that idea in 1 and see

1 + 2 + 3 - 4 = 2

but when you apply the same to equation 3

the negativity becomes larger and adding all the cubes will turn it negative.

so change the sign for 1 and 3 and 4

So the answer is -1,2,-3 and 4

Answer 7

@UnkleRhaukus

Answer 8

sorry squared and added is 1 + 4 + 9 + 16 = 30

Answer 9

Hey, @rajee_sam that is exactly the method i used !

Answer 10

So where is my medal and fanning????

Answer 11

what is my major prize???

Answer 12

\[\text{minor prize goes to}:\qquad\boxed{\stackrel{\color{goldenrod} {\ast\sim\star\sim\ast\sim\ast\sim\star\sim\ast}}{\color{purple}{\mathcal{RAJEE\_SAM}}}}\]

Answer 13

minor prize?? you said major prize

Answer 14

Ok first to solve I got it

Answer 15

major prize goes to the best method, i think there is another method , and i'd like to see it

Answer 16

OK If there are no other?? What is the deadline for entries?? You think yours is not the best?? LOL

Answer 17

well @UnkleRhaukus is there any other method

Answer 18

you tell me

Answer 19

This is a good method for these type of problems. Use the given system to solve for these equations:\[a+b+c+d=p_1\]\[ab+ac+ad+bc+bd+cd=p_2\]\[abc+bcd+acd+abd=p_3\]\[abcd=p_4\]Then treat a,b,c and d as the four solutions to a quartic polynomial:\[x^4-p_1x^3+p_2x^2-p_3x+p_4=0\]The solutions you get will be a, b, c, and d. Here is an example with a system of three equations:

http://openstudy.com/users/joemath314159#/updates/4e1002d70b8b56e555985df7