solutions of the equation x|x+1| +1=0 are
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x | x +1 | + 1= 0
this is modulus function
so, first of required to break of modulus .
for x > -1
x ( x +1 ) + 1 = 0
x² + x +1 = 0
here no real value so, x = zero solution
x< -1
x( - x -1) + 1 = 0
x² +x -1 = 0
x = { -1 ±√5}/2
but x < -1 so, x = (-1-√5)/2
is only solution .
this is modulus function
so, first of required to break of modulus .
for x > -1
x ( x +1 ) + 1 = 0
x² + x +1 = 0
here no real value so, x = zero solution
x< -1
x( - x -1) + 1 = 0
x² +x -1 = 0
x = { -1 ±√5}/2
but x < -1 so, x = (-1-√5)/2
is only solution .
Ramu111:
why x has no real values
in first case x> -1 this is only for real numbers not for imahinary
but x² + x +1 = 0 have both imaginary roots so,
contradiction
now you have understand
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