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Question Number 114420 by Aina Samuel Temidayo last updated on 19/Sep/20
Solution of ∣ x−1 ∣≥∣ x−3 ∣ is
Solutionof∣x−1∣⩾∣x−3∣is
Commented by bemath 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,∞)
shortcut⇔(x−1+x−3)(x−1−x+3)⩾0⇔(2x−4)(2)⩾0⇔x⩾2;x∈[2,∞)
Commented by Aina Samuel Temidayo last updated on 19/Sep/20
This is not well explanatory.
Thisisnotwellexplanatory.
Commented by bemath last updated on 19/Sep/20
no. your just square both sides
no.yourjustsquarebothsides
Commented by bobhans last updated on 19/Sep/20
mr bemath creative. santuyy
mrbemathcreative.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,+∞)
⇒∣x−1∣−∣x−3∣⩾0CASEI:whenx−1⩾0andx−3⩾0x⩾1andx⩾3Findingtheirintersections⇒x⩾3⇒∣x−1∣−∣x−3∣⩾0,x⩾3⇒x−1−(x−3)⩾0⇒x−1−x+3⩾0⇒2⩾0⇒x∈RRecallx⩾3⇒x⩾3ORCASEII:whenx−1<0andx−3<0⇒x<1andx<3Findingtheirintersections⇒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∈ϕORCaseIII:whenx−1<0andx−3⩾0⇒x<1andx⩾3Findingtheirintersections⇒x∈ϕORCaseIV:x−1⩾0andx−3<0⇒x⩾0andx<3Findingtheirintersections⇒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⩾2sincex∈[0,3)⇒x∈[2,3)CombiningCasesI,II,IIIandIV⇒x∈[2,+∞)
Commented by bemath last updated on 19/Sep/20
your answer very long like a road
youranswerverylonglikearoad
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;+∞)
Solvetheinequality∣x−1∣⩾∣x−3∣(1)Sincex−1⩾0⇔x⩾1,x−3⩾0⇔x⩾3Wehavethefollowingtablet:|x13∣x−1∣1−x0x−12x−1∣x−3∣3−x23−x0x−3|Fromabovetabletwehavei)Ifx⩽1then(1)⇔1−x⩾3−x⇔1⩾3⇒hasnorootsii)If1<x⩽3then(1)⇔x−1⩾3−x⇔2x⩾4⇔x⩾2wehavetherootsare2⩽x⩽3iii)Ifx>3then(1)⇔x−1⩾x−3⇔0.x⩾−2⇒∀x>3satisfyCombiningthreecaseswegettherootsofgiveninequalityarex∈[2;+∞)

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