The sum of all the real solutions of the equation (x+3) (6x2 + 28x + 30 ) = 5 - 2 (6x+10) (x2 + 6x + 9 ) is equal to :
- 1
- 4
- 0
- 2
Answer
(3) 0
Solution
6x2 + 28x + 30 = (x + 3)(6x + 10) and x2 + 6x + 9 = (x + 3)2
Let t = (x+3)(6x + 10). Then
LHS = 1 + t, RHS = 5 - 2 2t = 5 - 4t
t2 - 4t + 4 = 0 t = 2
6x + 10 = (x + 3)2
x2 = 1 x = 1
Check bases and arguments: x = 1: bases 4, 16; x = -1: bases 2, 4; all arguments positive ✓
Sum of solutions = 1 + (-1) = 0
Answer: (3)