Question #995d9

1 Answer
Manasvini Nittala
Oct 24, 2017

if in the given equation #(x^-3.x)+2=0# belongs to #RR# i.e. real numbers then the equation has no solution,if the equation belongs to complex numbers #CC# the solution is #(i)/sqrt2#

Explanation:

#x^-3.x+2=0rArr(x)/x^3+2=0rArr1/x^2+2=0rArr2x^2+1=0rArr2x^2=-1rArrx^2=(-1)/2rArrx=sqrt(-1)/sqrt2rArrx=i/sqrt2_#
hence this is how u solve it

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