Question

a+b+c=12
a^2 +b^2+ c^2 = 70
ab+bc+ca= 37
Find value a, b and c?

Answer 1

Given :   a² + b² + c² = 70,  a + b + c = 12, ab + bc + ca = 37

To find :   a, b and c

Solution:

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

=>  12² = 70 + 2(37)

=> 144 = 70 + 74

=> 144 = 144

Hence 3rd Equation can be found if 2 Equations are given

Hence these 3 Equations are basically 2 Equations

and 3 Variable a , b & c

We can not solve 3 variables with 2  Equations

There can be many possible solutions  

few are below :

a =  √37 i   b = -√37 i     c = 12

a =  -√37 i    b =  √37 i    c = 12

c=  √37 i    b = -√37 i     a = 12

c =  -√37 i    b =  √37 i     a = 12

a =  √37 i    c = -√37 i     b = 12

a =  -√37 i    c =  √37 i     b = 12

Verification  

a² + b² + c² =  -37 - 37 + 144 = 70   (i² = -1)

a + b + c = √37 i -√37 i  + 12 = 12

ab + bc + ca = (√37 i) * (-√37 i) + (√37 i) 12  +  (-√37 i) 12  = 37

Not Enough Details to find Unique Solutions

we can choose any value  then find others

let say a = 0

=> b + c = 12           bc = 37

=> x²  - 12x  + 37 = 0

=>  b & c   =    ( 12  ± √144 - 4 * 37 ) / 2    =  6 ±  i

a = 0   , b = 6 + i  , c = 6 - i

a + b + c = 12

a² + b² + c² = 0 + 36 - 1 + 12i + 36  - 1 - 12i   = 72

ab + bc + ca =  (6 + i)(6 - i)  = 36 - i² = 36 -(-1) = 37

Hence if we assume one value we can get  others

a = 1  => b + c = 11   & bc = 26

x² - 11x  + 26 = 0

b & c =  (11 ±  √17)/2

a = 2  => b + c = 10   & bc = 17

x² - 10x  + 17 = 0

b & c =  (10 ±  √32)/2   = 5 ± 2√2

and so on limit less solution

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