If 9a + 9b +9c = 81 and 4ab + 4bc + 4ca = 160, then what is the value of $6a^2 +6b^2 +6c^2 $ ?
Answer & explanation
Correct answer: option 4
If 9a + 9b +9c = 81
4ab + 4bc + 4ca = 160
what is the value of $6a^2 +6b^2 +6c^2 $ ?
If the number of equations are less than the number of variables then we can put the extra variables according to our choice =
So here two equations given and three variables are present so put c = 0
If 9a + 9b = 81
then, a + b = 9
4ab = 160
ab = 40
what is the value of $6a^2 +6b^2 $ ? = $6(a^2 + b^2) $ ?
( a + b )2 = a2 + b2 + 2ab
( 9 )2 = a2 + b2 + 2 × 40
81 = a2 + b2 + 80
a2 + b2 = 1
And, $6(a^2 + b^2) $ = 6(1) = 6