what is the equation of lowest degree with real coefficients if two of its roots are -1 and 1+i?
A. x^3+x^2+2=0
B. x^-x^2-2=0
C. x^3-x+2=0
D. x^3-x^2+2=0
e. none of the above
i don't get it is it (x+1) (1+i)
1)
x ^ 3 + a x ^ 2 + b x + c = ( x - x1 ) ( x - x2 ) ( x - x3 )
2)
For an equation with real coefficients, imaginary roots ( if any ) occur in conjugate pairs,
i.e. f ( x ) = 0 a polynomial equation with real coefficients an
n + i * m
an imaginary root of it , then
n - i * m
is also a root of it.
In this case :
x1 = - 1
x2 = 1 + i
x3 = 1 - i
Your equation becomes :
( x - 1) [ x - ( 1 + i ) ] [ x - ( 1 - i ) ]
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Remark:
[ x - ( 1 + i ) ] [ x - ( 1 - i ) ] = x ^ 2 - 2 x + 2
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( x - 1) [ x - ( 1 + i ) ] [ x - ( 1 - i ) ] =
( x - 1 ) ( x ^ 2 - 2 x + 2 ) =
x ^ 3 - x ^ 2 + 2
Answer : D
For any polynomial equation with rational coefficients, complex roots must occur as conjugate pairts, that is,
if one root is 1+i, there has to be another one 1 - i
so, including the root of -1, the equation could have been
(x+1)(x - (1+i) )(x - (1-i) ) = 0
(x+1)(x-1-i)(x-1+i) = 0
(x+1)(x^2 - 2x + 2) = 0
x^3 - 2x^2 + 2x + x^2 - 2x + 2 =0
x^3 - x^2 + 2 = 0
looks like D
Correction :
Your equation becomes :
( x + 1) [ x - ( 1 + i ) ] [ x - ( 1 - i ) ]
( x + 1) [ x - ( 1 + i ) ] [ x - ( 1 - i ) ] =
( x + 1 ) ( x ^ 2 - 2 x + 2 ) =
x ^ 3 - x ^ 2 + 2
Answer D
To find the equation of the lowest degree with real coefficients if two of its roots are -1 and 1+i, you need to remember that complex conjugate roots always occur in pairs.
So, if 1+i is one of the roots, then its conjugate, 1-i, must also be a root.
To find the equation, you can multiply the factors (x-a)(x-b)(x-c), where a, b, and c are the roots of the equation.
Since the given roots are -1, 1+i, and 1-i, the equation becomes:
(x - (-1))(x - (1+i))(x - (1-i)) = (x + 1)(x - 1 - i)(x - 1 + i)
Now, simplify the equation:
(x + 1)(x - 1 - i)(x - 1 + i) = (x + 1)((x - 1) - i)((x - 1) + i)
= (x + 1)((x - 1)^2 - i^2)
= (x + 1)(x^2 - 2x + 1 + 1)
= (x + 1)(x^2 - 2x + 2)
Expanding the equation further:
(x + 1)(x^2 - 2x + 2) = x(x^2 - 2x + 2) + 1(x^2 - 2x + 2)
= x^3 - 2x^2 + 2x + x^2 - 2x + 2
= x^3 - x^2 + 2
Therefore, the equation of the lowest degree with real coefficients, given that the roots are -1 and 1+i, is x^3 - x^2 + 2.
Looking at the answer choices, we find that none of the provided options (A, B, C, or D) match the equation x^3 - x^2 + 2. So, the correct answer is option E, none of the above.