If a square + b square + c square is equal to 14 then find the value of a b + BC + CA
littyissacpe8b60:
where is a+b+c= ?
Sorry but question is this only
we can't answer with this information only. I saw here someone put same question but with a+b+c = 6
with a+b+c = 4 also I saw in internet
Answers
Answered by
2
Given, a² + b² + c² = 14
we have to find value of ab + bc + ca
basically , we can get (ab + bc + ca) when one more date is given . I mean if values of (a + b + c) and a² + b² + c² both are given then we can easily find the value of ab + bc + ca. so, here we just use "Hit and trial method."
if we took , a = 1, b = 2 and c = 3
then, a² + b² + c² = 1² + 2² + 3² = 1 + 4 + 9 = 14
so, ab + bc + ca = 1 × 2 + 2 × 3 + 3 × 1
= 2 + 6 + 3 = 11
hence, ab + bc + ca = 11
we have to find value of ab + bc + ca
basically , we can get (ab + bc + ca) when one more date is given . I mean if values of (a + b + c) and a² + b² + c² both are given then we can easily find the value of ab + bc + ca. so, here we just use "Hit and trial method."
if we took , a = 1, b = 2 and c = 3
then, a² + b² + c² = 1² + 2² + 3² = 1 + 4 + 9 = 14
so, ab + bc + ca = 1 × 2 + 2 × 3 + 3 × 1
= 2 + 6 + 3 = 11
hence, ab + bc + ca = 11
Answered by
11
HEY DEAR ... ✌️
__________________________
Here's , Your Answer ...
=) Given ,
a² + b² + c² = 14
To find :- ab + bc + ca
Then , To find (ab + bc + ca) we have to find the values of (a + b + c) and a² + b² + c² .
So , "Hit and trial method" will be used .
if we assume , a = 1, b = 2 and c = 3
So , a² + b² + c²
= 1² + 2² + 3²
= 1 + 4 + 9
= 14
Also , ab + bc + ca
= 1 × 2 + 2 × 3 + 3 × 1
= 2 + 6 + 3
= 11
Hence , The value for ab + bc + ca = 11 .
_______________________
HOPE , IT HELPS ... ✌️
__________________________
Here's , Your Answer ...
=) Given ,
a² + b² + c² = 14
To find :- ab + bc + ca
Then , To find (ab + bc + ca) we have to find the values of (a + b + c) and a² + b² + c² .
So , "Hit and trial method" will be used .
if we assume , a = 1, b = 2 and c = 3
So , a² + b² + c²
= 1² + 2² + 3²
= 1 + 4 + 9
= 14
Also , ab + bc + ca
= 1 × 2 + 2 × 3 + 3 × 1
= 2 + 6 + 3
= 11
Hence , The value for ab + bc + ca = 11 .
_______________________
HOPE , IT HELPS ... ✌️
Similar questions