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1 solution
The equation can be rewritten as:
3 ( x + 1 ) 2 + 4 + 5 ( x + 1 ) 2 + 9 = − ( x + 1 ) 2 + 5
We have: 3 ( x + 1 ) 2 + 4 ≥ 2 and 5 ( x + 1 ) 2 + 9 ≥ 3 , so L H S ≥ 5 for all real x
Similarly, R H S = − ( x + 1 ) 2 + 5 ≤ 5 .
For the equation to occur, x + 1 = 0 , or x = − 1