Question

Find the values of m for which both roots of equation x² − mx + 1 = 0 are less than unity.

Answer 1

Let f (x) = x² − mx + 1. As both roots of f(x) = 0 are less than 1, we can take D ≥ 0, af(1) > 0 and −//2a < 1.

(a) Consider D ≥ 0 : (−m) 2 − 4⋅1⋅1 ≥ 0

⇒ (m + 2) (m − 2) ≥ 0

⇒ m ∈ (−∞, −2] ∪ [2, ∞) .....(1)

(b) Consider af(1) > 0:1⋅(1 − m + 1) > 0

⇒ m − 2 < 0

⇒ m < 2

⇒ m ∈ (−∞, 2)    .....(2)

(c)  Consider - b/2a < 1 ⇒ m/2 < 1

⇒ m < 2

⇒ m ∈ (−∞, 2) ....(3)

Hence, the values of m satisfying Eqs. (1), (2) and (3) at the same time are m ∈ (−∞, −2).