Given,
∣x−2∣+x(x−4)+2=0
⇒∣x−2∣+x−4x+2=0
⇒∣x−2∣+(x)2−4x+4=2 (Adding 2 on both sides)
⇒(x−2)2+∣x−2∣=2
⇒t2+t−2=0; ∣x−2∣=t;(t>0)
⇒t=1or t=−2 ; Rejected because t>0
⇒∣x−2∣=1
⇒x=3 or x=1
⇒x=9 or x=1
Hence, the required sum =10.
JEE Main 2019 — Mathematics Algebra
The sum of the solutions of the equation ∣x−2∣+x(x−4)+2=0,(x>0) is equal to
Held on 8 Apr 2019 · Verified 6 Jul 2026.
10
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12
4
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