Does anybody know how he got x^2+4x+5?

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  • 1 month ago

    (x - (-2 +i))(x-(-2 -i))

    = ((x+2) - i)((x+2) + i)

    = (x+2)^2 - i^2

    = x^2 + 4x + 4 + 1

    = x^2 + 4x + 5

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  • 1 month ago

    If an equation has a complex root like x=-2+i, it must

    have its conjugate x=-2-i as a root. So,

    (x+2-i)(x+2+i)=(x+2)^2-i^2=x^2+4x+5 is a factor of the given equation f(x)=0. Thus,

    f(x)=(x+2-i)(x+2+i)(x^2+6x+9)=0

    =>

    (x+2-i)(x+2+i)(x+3)^2=0

    =>

    x=-2+i

    x=-2-i

    x=-3 (repeated roots)

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  • Philip
    Lv 6
    1 month ago

    Since x = -2-i and x = -2+i are both zeros of 4th degree

    polynomial function f(x) = x^4 +10x^3 +38x^2 +66x +45,

    [x-(-2-i)][x-(-2+i)]=(x+2+i)(x+2-i)=(x+2)^2 -i^2=x^2+4x+5 is

    a factor of f(x). That's how he got it! Dividing f(x) by

    x^2+4x+5 gave the quadratic g(x) = x^2+6x+9 = (x+3)^2.

    The 2 zeros of g(x) are x = -3{multiplicity 2} which are the

    final 2 zeros of f(x). 

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  • 1 month ago

    find roots for : x^4 +10x^3 +38x^2 +66x +45 = 0, ------------------- [1]

    Roots are: x1 = -2+i, x2 = -2-i, x3 =?, and x4 = ?

    x1 * x2 = (x+2-i)(x+2+i) = x^2 +x(2-i +2+i) + (2-i)(2+i) = x^2+4x +5

    So,

    x^4 +10x^3 +38x^2 +66x +45 = (x^2+4x+5)(x^2 +bx +c) = 0,

    Or,

    x^2(x^2+bx+c) +4x(x^2+bx+c) +5(x^2+bx+c) = 0,

    Or,

    x^4 +x^3(b+4) +x^2(c+4b+5) +x(4c+5b) +5c = 0, ..................[2]

    Comparing [1] & [2],

    b+4 = 10, 4b+c+5 = 38,  4c+5b = 66, and 5c = 45

    b = 6, 24+c+5 = 38, 4c+5b = 66, and c + 9,

    Or,

    b=6, c= 9, 

    Hence,

    x^4 +10x^3 +38x^2 +66x +45 = (x^2+4x+5)(x^2 +6x +9) = 0,

    For,

    x^2+4x+5 = 0, x= -2+i, & x = -2-i,

    and for,

    x^2 +6x +9 = 0, x = -3, and x = -3

    The roots are:

    x= -2+i,  x = -2-i,   x = -3, and x = -3. ------------------ Anwr

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  • sepia
    Lv 7
    1 month ago

    x^4 + 10x^3 + 38x^2 + 66x + 45 = 0

    (x + 3)^2 (x^2 + 4 x + 5) = 0

    Real solution:

    x = -3

    Complex solutions:

    x = -2 - i

    x = -2 + i

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  • 1 month ago

    We are told one solution and we can infer the conjugate:

    x = -2 + i

    x = -2 - i

    Rewrite those as expressions equal to 0:

    x + 2 - i = 0

    x + 2 + i = 0

    Now group x+2:

    (x + 2) - i = 0

    (x + 2) + i = 0

    Finally multiply them together:

    [(x + 2) - i][(x + 2) + i]

    It's a difference of squares:

    (x + 2)² - i²

    = x² + 4x + 4 - (-1)

    = x² + 4x + 5

    • ...Show all comments
    • Puzzling
      Lv 7
      1 month agoReport

      I was explaining why the person wrote down the conjugate pair. It was not given originally. 

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  • Alvin
    Lv 4
    1 month ago

    yes, 

    root of x = -2 + i 

    and x = -2 -i  

    means that 

    (x+ 2 -i)(x+ 2 +i) are factors 

    use distributive rule to 

    =x^2 + 2x +ix   + 2x + 4 + 2i  -ix -2i  - (i^2)  = 0 

    = x^2   + (2x + 2x -ix + ix )   + 4  - (-1)  = 0 

    = x^2 + 4x  + 4 + 1  

    =x^2 + 4x  + 5   (that's how

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  • 1 month ago

     x^4 + 10x^3 + 38x^2 + 66x + 45 = 0

    (x + 3)^2 (x^2 + 4 x + 5) = 0

    Real solution:

    x = -3

    Complex solutions:

    x = -2 - i

    x = -2 + i

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  • ted s
    Lv 7
    1 month ago

    you are told that - 2 ± i are roots ===> (x - (-2 ))² + 1 is a factor = x² + 4x + 5

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  • 1 month ago

    change the two solutions into factors then multiply the two factors together

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