Question

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find the value for which both roots of x²-mx+1=0 less than unity.

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Answer 1

Answer:

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)

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Answer 2

Answer:

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