A) 4
B) 2
C) 0
D) None of these
Correct Answer: B
Solution :
\[a{{x}^{2}}+2b|x|-c=0\] \[\Rightarrow \] \[a|x{{|}^{2}}+2b|x|-c=0\] \[\therefore \]\[|x|\,\,=\frac{-2b\pm \sqrt{4{{b}^{2}}+4ac}}{2}=-b\pm \sqrt{{{b}^{2}}+ac}\] Since,\[a,\,\,b,\,\,c\]are positive. So, \[|x|=-b+\sqrt{{{b}^{2}}+ac}\] \[\therefore \]\[x\]has two real values, neglecting \[|x|=-b-\sqrt{{{b}^{2}}+ac}\], as\[|x|\,\,>0\]You need to login to perform this action.
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