Question

If a+b+c=0 and a^2+b^2+c^2=22, what is a^4+b^4+c^4?

Answer 1

you could assume that a = b+c
so if a + b + c = 0 = true
then
a^2+b^2+c^2=22
so as a = b+c
(b+c)^2 +b^2+c^2=22

Answer 2

then

Answer 3

(b+c)^2 +b^2+c^2=22
b^2 +2bc + c^2 +b^2 + c^2 = 22
2b^2+ 2c^2 + 2bc = 22

Answer 4

Then?

Answer 5

then reduce, so b = something + c
sub back into eqn 1
solve simultaneous equations

Answer 6

help me for that

Answer 7

factor this: 2b^2+ 2c^2 + 2bc = 22
in terms of b

Answer 8

first: divide both sides by 2

Answer 9

then try quadratic solution after rearranging for = 0...?

Answer 10

2b^2+ 2c^2 + 2bc = 22
so 2b^2+ 2c^2 + 2bc -22 = 0
so b^2+ c^2 + bc -11 = 0
use quadratic solution
b = .5 + or - sqrt ((44 - 3c^2) - c)

Answer 11

sub back in, solve for c
so c = sqrt (11/3)

therefore b = -2 (sqrt (11/3))

Answer 12

so what is the answer

Answer 13

so a must = c = sqrt (11/3)
(as a+b+c = 0, in this case a+c = b as we solved for b first)

Answer 14

now just put each to the power of 4 to find ur answer

Answer 15

answer?

Answer 16

answer is a^4+b^4+c^4
where:
a = sqrt (11/3)
b = -2 (sqrt (11/3))
c = sqrt (11/3)

remember that sqrt a = a^(1/2)

Answer 17

what is the value of a^4+b^4+c^4

Answer 18

u work it out, i've given you the values of a and b and c

Answer 19

im here to help you do your work, not do it for you
how bout u give it a try, and i'll tell u if you're right?

Answer 20

let me try

Answer 21

1210/9

Answer 22

no.

Answer 23

try approximating the fractions into just numbers, less confusion that way
what's 11/3 =

Answer 24

3.66

Answer 25

then take the sqr root of that

Answer 26

0.6

Answer 27

no, 1.915

Answer 28

oh i see
then?

Answer 29

a = 1.915
so a^4 = 1.915 x 1.915 x 1.915 x 1.915 =???

Answer 30

13.44

Answer 31

perfect
now c = 1.915
so c^4 =???

Answer 32

13.44

Answer 33

spot on,
now b = -2 x sqrt 11/3
=-2 x 1.915
=?

Answer 34

-3.830

Answer 35

close enough, i got -3.8297
so b = -3.830
therefore b^4 = -3.830 x -3.830 x -3.830 x -3.830 = ???

Answer 36

215.17

Answer 37

yep, now add them together

Answer 38

thank you

Answer 39

welcome
what was ur final answer?

Answer 40

242.( )( )

Answer 41

yep, the . is only there because of rounding and approximating, whole answer is 242 even

Answer 42

@Jack1 - how can you assume a = b + c in your first step?

Answer 43

sorry, a = -b-c

Answer 44

if a + b + c = 0
... my bad

Answer 45

ok - that makes more sense

Answer 46

and i ended up choosing b to equal (-c + -a) instead anyway

Answer 47

as my calcs ended up with:
a = sqrt (11/3)
b = -2 (sqrt (11/3))
c = sqrt (11/3)

Answer 48

here is an alternative method to solve this. given:\[a+b+c=0\tag{1}\]\[a^2+b^2+c^2=22\tag{2}\]then:\[\begin{align}
(a+b+c)^2&=a^2+b^2+c^2+2(ab+bc+ca)=22+2(ab+bc+ca)\\
\therefore 0&=22+2(ab+bc+ca)\\
\therefore ab+bc+ca&=-11\tag{3}\\
(ab+bc+ca)^2&=a^2b^2+b^2c^2+c^2a^2+2(ab^2c+a^2bc+abc^2)\\
&=a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)\\
\therefore (-11)^2&=a^2b^2+b^2c^2+c^2a^2+0\\
\therefore a^2b^2+b^2c^2+c^2a^2&=121\tag{4}
\end{align}\]then finally we can get:\[\begin{align}
(a^2+b^2+c^2)^2&=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)\\
\therefore (22)^2&=a^4+b^4+c^4+2(121)\\
\therefore a^4+b^4+c^4&=22^2-242=484-242=242\\
\end{align}\]

Answer 49

much cleaner