Sum of the squares of all integral values of a for which the inequality x 2 + a x + a 2 + 6 a < 0 is satisfied for all x ∈ 1 , ⁡ 2 must be equal to

Step-by-step solution

Let, f ⁡ x = x 2 + a x + a 2 + 6 a ∴ f ( 1 ) ≤ 0 ⇒ a 2 + 7 a + 1 < 0 or - 7 - 3 5 2 < a < - 7 + 3 5 2 ....(i) f ( 2 ) ≤ 0 ⇒ a 2 + 8 a + 4 < 0 or - 4 - 2 3 < a < - 4 + 2 3 ....(ii) and D > 0 ⇒ a 2 - 4 · 1 a 2 + 6 a > 0 ⇒ a 2 + 8 a < 0 or - 8 < a < 0 ....(iii) From Eqs. (i), (ii) and (iii), we get, - 7 - 3 5 2 ≤ a ≤ - 4 + 2 3 Hence, integral values of a are - 6 , ⁡ - 5 , ⁡ - 4 , ⁡ - 3 , ⁡ - 2 , ⁡ - 1 Required Sum = - 6 2 + - 5 2 + - 4 2 + - 3 2 + - 2 2 + - 1 2 = 91 .