Case-I
x≤5
(x+1)2−(x−5)=427
(x+1)2−(x+1)−43=0
x+1=23,−21
x=21,−23
Case-II
x>5
(x+1)+(x−5)=427
(x+1)2+(x+1)−451=0
x=2−1±52 (rejected as x>5)
So, the equation have two real root.
The number of the real roots of the equation (x+1)2+∣x−5∣=427 is ________.
Held on 24 Feb 2021 · Verified 6 Jul 2026.
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