if a square + b square + c square is equal to 20 and a + b + C is equal to zero then the value of AB + BC + CA is
Answers
Answer:
Given :-
a² + b² + c² = 20
a+b+c = 0
Using this formula :
(a+b+c)² = a² + b² + c² + 2(ab + bc + ca)
So,
(0)² = 20 + 2(ab + bc + ca)
=> 0 = 20 + 2(ab + bc + ca)
=> -20 = 2(ab + bc + ca)
=> -20/2 = (ab + bc + ca)
=> -10 = (ab + bc + ca)
Hence,
ab+ bc+ca = -10
Answer:
-10
Explanation :
Take a + b + c = 0 and square both sides of the equation..
(a + b + c)^2 = 0^2
(a + b + c)(a + b + c) = 0
Apply the distributive property: a (b + c) = ab + ac...multiply every term in one trinomial and multiply by the other trinomial....and remember that when you multiply expressions with the same base, you add their exponents.
(a + b + c)(a + b + c) = 0
a (a + b + c) + b (a + b + c) + c (a + b + c) = 0
a^(1+1) + ab + ac + ba + b^(1+1) + bc + ca + cb + c^(1+1) = 0
a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2 = 0
Group the a^2 + b^2 + c^2 together and gather the like terms together (I am going to rearrange these terms such that like terms, terms with the same variable(s) raised to the same power, are together).
a^2 + b^2 + c^2 + ab + ba + ac + ca + bc + cb
Combine like terms (add the coefficients of the like terms in this case) and we are given that: a^2 + b^2 + c^2 = 20..so..make the substitution.
20 + (1 + 1)ab + (1 + 1)ac + (1+1)bc = 0
20 + 2ab + 2ac + 2bc = 0
Subtract 20 to both sides of the equation...
20 + 2ab + 2ac + 2bc - 20 = 0 - 20
2ab + 2ac + 2bc = -20
Factor out the greatest common factor of 2 out of the left hand side..
2 (ab + ac + bc) = -20
Divide both sides of the equation by 2 to isolate the expression...
2 (ab + ac + bc)/2 = -20/2
ab + ac + bc = -10