A+B+C=12 and A^2+B^2+C^2=70 and
AB+BC+AC=37 then find the value of A ,B ,C
Answers
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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