Question

Could I please have some assistance with this question? http://i.imgur.com/rbg1uxR.jpg

Answer 1

Thats actually a very interesting equaiton.

Answer 2

Yes it is, I'm learning quadratic equations right now, and this was one of the harder problems.

Answer 3

Could we move the A, B and C values (not squared) to the RHS. Then take the square as the common value?

Answer 4

not sure how this would help but
\(\Large (a+2)^2 + (b-6)^2 + (c+5)^2 = 0\)

because 65 can be split as 4 +36+ 25

Answer 5

Actually that's a very good idea!

Answer 6

but i don't know that helps to solve the equation...

Answer 7

\[\large{a^2+b^2+c^2+4a-12b+10c=-65}\]
\[\large{a^2+b^2+c^2=-65-4a+12b-10c}\]
\[\large{(a+b+c)^2=-65-4a+12b-10c}\]Then, go to 2015, maybe?

Answer 8

haha and how do i increase it to that power?

Answer 9

go by even powers , first , i would try

Answer 10

(BTW, is what I did even help?)

Answer 11

yeah it helps a lot actually

Answer 12

\(a^2+b^2+c^2 \ne (a+b+c)^2\)

Answer 13

Oh, damnit :(

Answer 14

Is there anyway we could take the square as common?

Answer 15

what does that mean? @Ahsome

Answer 16

To remove the square from the LHS.

Answer 17

$$
\large{

(a+b+c)^2=-65-4a+12b-10c\\
\\ \therefore\\
(a+b+c)^4\\
=(-65-4a+12b-10c)^2\\
=(4a-12b+10c+65)^2
\\ \therefore\\
(a+b+c)^4=(4a-12b+10c+65)^2
\\ \therefore \\
(a+b+c)^8=(4a-12b+10c+65)^4
\\
etc.
}\\


$$

Answer 18

but a^2 + b^2 + c^3 doesn't equal the (a+b+c)^2

Answer 19

Yeah. We established that after @perl started to solve :(

Answer 20

what the guy said at the start was right, i found the solution. 65 can be expressed as 4, 25 and 36.

Answer 21

you then divide them into three trinomials.

Answer 22

you got it using my hint? :O
please enlighten us too :)

Answer 23

@perl
\(a^2+b^2+c^2 \ne (a+b+c)^2\)

Answer 24

i dont think i claimed that
unless i made a mistake somewhere

Answer 25

you started with (a+b+c)^2 itself... instead of starting with initial equation...

Answer 26

oh Ahsome claimed it, and i might have worked off that

Answer 27

*egad*

Answer 28

Yup, whoops, sorry @perl

Answer 29

@Sepeario do you have the answer, i'm not sure if you have a solution or not

Answer 30

Don't worry, I figured it out, but thanks for the help guys!

Answer 31

How????

Answer 32

yes please show

Answer 33

oh wow!

Answer 34

*facepalm* :P

Answer 35

yes that looks too easy :)

Answer 36

but hartnn did the hard work. the tricky part was finding a suitable factoring of the original expression

Answer 37

if a^2+b^2+c^2 = 0
then a = b= c = 0 is a must :D

Answer 38

quick question how do we find the values of a b and c?

Answer 39

solve that simultaneously

Answer 40

you did find it
\(\Large (a+2)^2 + (b-6)^2 + (c+5)^2 = 0 \\ \Large
a+2 = 0; b-6 = 0 , c+5 = 0 \\ \Large a= -2, b = 6, c = -5\)

Answer 41

oh right lol

Answer 42

hartn, so you basically completed the square

Answer 43

for the sphere

Answer 44

actually its a degenerate sphere (a single point)

Answer 45

thats right

Answer 46

treating a,b,c as variables

Answer 47

nice