Question 30217
Explain why x to the 4th+2xsquared+4 has no real root 
X^4+2X^2+4={(X^2)^2+2(X^2)(1)+(1)^2}-(1)^2+4 = (X^2+1)^2+3...OR...(X^2+1)^2=-3
WE FIND THT THE SQUARE OF A NUMBER X^+1 IS NEGATIVE WHICH HAS NO REAL SOLUTION.(X^2+1)=i*SQRT.3
HENCE 
X^2+1= i*SQRT.3
X^2=-1+i*SQRT.3 = A COMPLEX NUMBER..HENCE ALL FOUR ROOTS OF X ARE COMPLEX/IMAGINARY NUMBERS.THAT IS THERE ARE NO REAL SOLUTIONS FOR X.
IT IS POSSIBLE HERE SINCE THE TWO PAIRS OF COMPLEX ROOTS ARE CONJUGATE WITH THE RESULT THAT THE FUNCTIONS OF THEIR  SUM/PRODUCT IS REAL.SINCE THE COEFFICIENTS OF VARIABLE IN THE POLYNOMIAL ARE FUNCTIONS OF  SUM/PRODUCT OF THE ROOTS,IT IS NECESSARY THAT THE ROOTS SHOULD BE REAL OR CONJUGATE COMPLEX NUMBERS FOR THIS TO HAPPEN IN CASE OF A POLYNOMIAL WITH REAL COEFFICIENTS.IN THIS CASE WE HAVE 
IF A,A',B,B' ARE ROOTS (A AND A' ARE CONJUGATE ...B AND B' ARE CONJUGATE).. THEN
A+A'+B+B'=0....POSSIBLE SINCE A+A' IS REAL AND B+B' IS REAL
AA'+AB+AB'+A'B+A'B'+BB'=2...POSSIBLE SINCE AA' AND BB' ARE REAL AND 
AB+AB'+A'B+A'B'=A(B+B')+A'(B+B')=(A+A')(B+B')=REAL*REAL=REAL 
.....ETC...

while every polynomial function of degree 3 has at least 1 real root
IN CASE OF 3 RD. DEGREE POLYNOMIAL ,WE HAVE 3 ROOTS..SO IF THEY ARE COMPLEX THEN AS WE SHOWED ABOVE FOR POLYNOMIAL WITH REAL COEFFICIENTS ..COMPLEX ROOTS OCCUR IN PAIRS OF CONJUGATE NUMBERS ONLY.SO IN 3 ROOTS 2 COULD BE COMPLEX CONJUGATES ,BUT THE 3 RD. HAS TO BE REAL IF THE COEFFICIENTS ARE REAL.