x 2 − 4 x > − 5 0" class="fm-inline"> ⇒ x 2 − 4 x + 5 > 0 , which is true ∀ x ∈ R ⇒ x ∈ ( − ∞ , ∞ ) .......(ii) From Eqs. (i) and (ii), x ∈ ( − 1 , 5 )" >
We have, |x2−4x|<5 ∴−5<x2−4x<5 Case I x2−4x<5 ⇒x2−4x−5<0 ⇒(x−5)(x+1)<0 ⇒x∈(−1,5) .......(i) Case IIx2−4x>−5 ⇒x2−4x+5>0, which is true ∀x∈R ⇒x∈(−∞,∞) .......(ii) From Eqs. (i) and (ii), x∈(−1,5)
