The value of x; ∀x ∈ R which satisfy the equation x - 1 x 2 - 4 x + 3 + 2 x 2 + 3 x - 5 = 0 is

Step-by-step solution

x - 1 x 2 - 4 x + 3 + 2 x 2 + 3 x - 5 = 0 Case I: when x 2 - 4 x + 3 ≥ 0 , then x ∈ ( - ∞ , 1 ] ∪ [ 3 , ∞ ) x - 1 x 2 - 3 x - x + 3 + 2 x 2 + 5 x - 2 x - 5 = 0 x - 1 x 2 - 3 x - x + 3 + 2 x + 5 = 0 x - 1 x 2 - 2 x + 8 = 0 (the value of quadratic can't be zero since its D < 0 ) ∴ x - 1 = 0 ⇒ x = 1 Case II: when x 2 - 4 x + 3 < 0 , then x ∈ 1 , 3 x - 1 - x 2 + 4 x - 3 + 2 x + 5 = 0 x - 1 ≠ 0 ∴ - x 2 + 6 x + 2 = 0 x 2 - 6 x - 2 = 0 x = 3 ± 1