prove that imaginary roots always occur in conjugate pairs
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It follows from the present theorem and the fundamental theorem of algebra that if the degree of a real polynomial is odd, it must have at least one real root. Since non-realcomplex roots come in conjugate pairs, there are an even number of them; Therefore some of them must be real
I hope bro this will help then it's upto you to choose as brainliest answer
I hope bro this will help then it's upto you to choose as brainliest answer
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we Can prove this by using polynomial concept .
we take 1st simple quadratic polynomial
e.g x² + 1 = 0 ( I take this one becoz this give imaginary roots )
x² + 1 = 0
x² = -1
x = ±√(-1) = ±i
hence, roots of x² + 1 are +i and -i
+i is conjugate pair of -i and vice -versa
again we take a simple cubic polynomial e.g x³ + 1 =0
x³ + 1 = 0
(x + 1)(x² + x + 1) = 0
x = -1 and x² - x +1 = 0
x² - x + 1 = 0
this polynomial given imaginary roots becoz D = 1 -4 = -3< 0
use quadratic formula ,
x = { 1 ±√3i }/2
hence, roots of x³ +1 = 0 are -1 , ( 1 +√3i)/2 , and ( 1 -√3i)/2
here ( 1+√3i)/2 is conjugate of ( 1 -√3i)/2
we saw , even degree and odd degree both taken conjugate pair of imaginary roots .
hence, if any polynomial have one root imaginary it means it has must one more root imaginary which is conjugate of known imaginary roots .
[ Note :- why imaginary roots exist in conjugate form ???
we know definition of polynomial all coefficient of dependent terms are real numbers so, sum of roots and products of roots should be real , How this is possible?? . this is possible only when they conjugate pair ]
Lemme one example :-
two imaginary roots a - ib and a + ib are roots of Px² + Qx + R = 0
where P, Q and R are real numbers
sum of roots = -Q/P
RHS = Q/P is real number so , LHS must be real number
see this
sum of roots = a- ib + a +ib = 2a is real number
also products of roots = R/P
R/P is real number
( a -ib)(a + ib) =( a² + b² ) is real number ]
we take 1st simple quadratic polynomial
e.g x² + 1 = 0 ( I take this one becoz this give imaginary roots )
x² + 1 = 0
x² = -1
x = ±√(-1) = ±i
hence, roots of x² + 1 are +i and -i
+i is conjugate pair of -i and vice -versa
again we take a simple cubic polynomial e.g x³ + 1 =0
x³ + 1 = 0
(x + 1)(x² + x + 1) = 0
x = -1 and x² - x +1 = 0
x² - x + 1 = 0
this polynomial given imaginary roots becoz D = 1 -4 = -3< 0
use quadratic formula ,
x = { 1 ±√3i }/2
hence, roots of x³ +1 = 0 are -1 , ( 1 +√3i)/2 , and ( 1 -√3i)/2
here ( 1+√3i)/2 is conjugate of ( 1 -√3i)/2
we saw , even degree and odd degree both taken conjugate pair of imaginary roots .
hence, if any polynomial have one root imaginary it means it has must one more root imaginary which is conjugate of known imaginary roots .
[ Note :- why imaginary roots exist in conjugate form ???
we know definition of polynomial all coefficient of dependent terms are real numbers so, sum of roots and products of roots should be real , How this is possible?? . this is possible only when they conjugate pair ]
Lemme one example :-
two imaginary roots a - ib and a + ib are roots of Px² + Qx + R = 0
where P, Q and R are real numbers
sum of roots = -Q/P
RHS = Q/P is real number so , LHS must be real number
see this
sum of roots = a- ib + a +ib = 2a is real number
also products of roots = R/P
R/P is real number
( a -ib)(a + ib) =( a² + b² ) is real number ]
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