SOLUTION: Explain why x to the 4th+2xsquared+4 has no real root while every polynomial function of degree 3 has at least 1 real root

Question 30217: Explain why x to the 4th+2xsquared+4 has no real root while every polynomial function of degree 3 has at least 1 real root
Found 2 solutions by Fermat, venugopalramana:
Answer by Fermat(136) (Show Source): You can put this solution on YOUR website!
x^4 + 2x^2 + 4
let u = x^2
then
u^2 + 2u + 4 = 0
when using the quadratic formula, , in order to get real roots, the discriminant, , must be positive.
In our quadratic eqn above,
a = 1, b = 2, c = 4
b�-4ac = 4 - 4*1*4 = 4 - 16 = -12
i.e. the discriminant is negative, hence the eqn has no real solutions - only complex ones.
Since u is complex, then so also is x.
Fact: all polynomials of the nth degree have n solutions.
Fact: when complex solutions (to a polynomial) occur, they occur as conjugate pairs.
A cubic eqn is of the 3rd degree, hence has three solutions. It has either zero complex solutions or two complex solutions. Hence it has at least one real solution.
N.B. a+ib and a-ib are conjugate pairs. (change of sign for the imaginary component)
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
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.

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