Learner 1: x = 1, because if x2- 2x = -1, then x² = 2x – 1 and x = V2x - 1. x can't be 0 because we get 0 = v-1. x can't be negative because we get the square root of a negative. x = 1 works because we get 1 = 1 and no other number bigger than 1 works %3D Learner 2: x = 1, because if x² – 2x =-1, then x(x – 2) = –1 and so x = -1 or x – 2 = -1, which leaves us with x = 1 (because x = -1 does not hold %3D !! true) Learner 3: x = 1, because if x² – 2x = -1, then x² – 2x+1 = 0 and this factorises to get (x – 1)(x – 1) = 0; so x = 1 Learner 4: x = 1. I drew the graphs y = 1 and y = x² – 2x. They intersect in only one place, at x = 1. - Learner 5: x = 1. | substituted a range of values for x in the equation and 1 is the only one that works.

Learner 1: x = 1, because if x2- 2x = -1, then x² = 2x – 1 and x = V2x - 1. x can't be 0 because we get 0 = v-1. x can't be negative because we get the square root of a negative. x = 1 works because we get 1 = 1 and no other number bigger than 1 works %3D Learner 2: x = 1, because if x² – 2x =-1, then x(x – 2) = –1 and so x = -1 or x – 2 = -1, which leaves us with x = 1 (because x = -1 does not hold %3D !! true) Learner 3: x = 1, because if x² – 2x = -1, then x² – 2x+1 = 0 and this factorises to get (x – 1)(x – 1) = 0; so x = 1 Learner 4: x = 1. I drew the graphs y = 1 and y = x² – 2x. They intersect in only one place, at x = 1. - Learner 5: x = 1. | substituted a range of values for x in the equation and 1 is the only one that works.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section: Chapter Questions

Problem 63RE

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