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WBJEEWBJEE 2010Complex Numbers and Quadratic Equations
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Solution:
Case I : When x>0,∣x∣=x ∴x2+15∣x∣+14=0 becomes x2+15x+14=0 ⇒x2+x+14x+14=0 ⇒x=−1,−14...(i)
Case II : When x<0,∣x∣=−x ∴x2+15∣x∣+14=0 becomes ⇒x2−15x+14=0 ⇒x2−x−14x+14=0 ⇒x(x−1)−14(x−1)=0 ⇒(x−1)(x−14)=0
(ii) ⇒x=1,14....(ii)
From Eqs. (i) and (ii) roots of given equation are not same. Therefore the given equation has no solution.