1928
226
Rajasthan PETRajasthan PET 2005
Report Error
Solution:
Given, equation is x2+5∣x∣+4=0
If x>0, then ∣x∣=x ∴x2+5x+4=0 ⇒x2+4x+x+4=0 ⇒(x+4)(x+1)=0 ⇒x=−4,−1 If x<0, then ∣x∣=−x ∴x2−5x+4=0 ⇒x2−4x−x+4=0 ⇒(x−4)(x−1)=0⇒x=4,1
Hence, no real roots exist.