The set of all values of k > - 1 , for which the equation 3 x 2 + 4 x + 3 2 - ( k + 1 ) 3 x 2 + 4 x + 3 3 x 2 + 4 x + 2 + k 3 x 2 + 4 x + 2 2 = 0 has real roots, is:

Step-by-step solution

Let t = 3 x 2 + 4 x + 2 ⇒   t + 1 2 - k + 1 t t + 1 + k t 2 = 0 ⇒ t 2 + 2 t + 1 - ( k + 1 ) t 2 - ( k + 1 ) t + k t 2 = 0 ⇒ 2 t + 1 - k t - t = 0 ⇒   t 1 - k = - 1 Replacing the value of t again 3 x 2 + 4 x + 2 = t = 1 k - 1 ⇒ 3 x 2 + 4 x + 2 k - 3 k - 1 = 0 For real roots, D ≥ 0 ⇒ 16 - 4 × 3 × 2 k - 3 k - 1 ≥ 0 ⇒ 6 k - 9 k - 1 - 4 ≤ 0 ⇒   2 k - 5 k - 1 ≤ 0   ⇒ k ∈ ( 1 , 5 2 ]