Let |a| 2 - |α|x-|β|=0. If|α| |a| - 1 = 0

|α| = a + b- (β) = abHence, a is negative b is positiveNow |α| ⇒ a

Math miscellaneousHard

Question

Let |a| < |b| and a, b are the roots of the equation x² - |α|x-|β|=0. If|α| < b - 1 then the equation log_|a| - 1 = 0

A.At least one root lying between (- ∞, a)

B.at least one root lying between (b, ∞)

C.has a negative root

D.has a positive root

|α| = a + b
- (β) = ab
Hence, a is negative b is positive
Now |α| < b - 1
⇒ a < - 1

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