Question Number 114420 by Aina Samuel Temidayo last updated on 19/Sep/20

Solution of  ∣ x−1 ∣≥∣ x−3 ∣ is

Commented bybemath last updated on 19/Sep/20

short cut  ⇔ (x−1+x−3)(x−1−x+3)≥0  ⇔(2x−4)(2)≥ 0  ⇔ x ≥ 2 ; x∈ [2,∞)

Commented byAina Samuel Temidayo last updated on 19/Sep/20

This is not well explanatory.

Commented bybemath last updated on 19/Sep/20

no. your just square both sides

Commented bybobhans last updated on 19/Sep/20

mr bemath creative. santuyy

Answered by Aina Samuel Temidayo last updated on 19/Sep/20

⇒∣x−1∣ −∣x−3∣≥0  CASE I:  when x−1≥0 and x−3≥0  x≥1 and x≥3  Finding their intersections  ⇒x≥3  ⇒ ∣x−1∣−∣x−3∣≥0 , x≥3  ⇒ x−1−(x−3)≥0  ⇒ x−1−x+3≥0  ⇒ 2≥0  ⇒ x∈R  Recall x≥3 ⇒ x≥3                              OR  CASE II:  when x−1<0 and x−3<0  ⇒ x<1 and x<3  Finding their intersections  ⇒x<1  ⇒ ∣x−1∣−∣x−3∣≥0  = −(x−1)−(−(x−3))≥0  ⇒ −x+1−(−x+3)≥0  ⇒−x+1+x−3≥0  ⇒−2≥0  ⇒ x∈φ                                OR  Case III:  when x−1<0 and x−3≥0  ⇒ x<1 and x≥3   Finding their intersections  ⇒ x∈φ                                  OR  Case IV:  x−1≥0 and x−3<0  ⇒ x≥0 and x<3  Finding their intersections  ⇒ x∈ [0,3)  ⇒ ∣x−1∣−∣x−3∣≥0  = x−1−(−(x−3))≥0  ⇒ x−1−(−x+3)≥0  ⇒ x−1+x−3≥0  ⇒ 2x≥4  ⇒ x≥2  since x∈[0,3)  ⇒x∈[2,3)    Combining Cases I,II,III and IV  ⇒ x∈  [2,+∞)

Commented bybemath last updated on 19/Sep/20

your answer very long like a road

Answered by 1549442205PVT last updated on 19/Sep/20

Solve the inequality ∣x−1∣≥∣x−3∣(1)   Since x−1≥0⇔x≥1,x−3≥0⇔x≥3  We have the following tablet:   determinant ((x,,1,,3,),((∣x−1∣),(1−x),0,(x−1),2,(x−1)),((∣x−3∣),(3−x),2,(3−x),0,(x−3)))  From above tablet we have  i)If x≤1 then  (1)⇔1−x≥3−x⇔1≥3⇒has no roots  ii)If 1<x≤3 then   (1)⇔x−1≥3−x⇔2x≥4⇔x≥2  we have the roots are 2≤x≤3  iii)If x>3 then   (1)⇔x−1≥x−3  ⇔0.x≥−2 ⇒∀x>3 satisfy   Combining three cases we get the  roots of given inequality are  x∈[2;+∞)