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2014-09-19T12:32:56+05:30
 x^{2} +8 = -1
or, x^{2}  = -9

we know that the square of any number will be +ve.
but given that -ve. it means number will be imaginary  
so,   x=  \sqrt{-9} =  \sqrt{ 3^{2} *(-1)} =3i
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just say x does not have any value ranging from negative infinity to positive infinity
yes offcourse
2014-09-19T12:52:46+05:30
X^2+8=-1
adding -8 on both sides
x^2+8-8=-1-8
x^2=-9
x=√-9
x=3i
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