finding polynomial equation from given complex conjugate pair roots

[tjackson8684]

anyone help me out?

we're given the complex roots are 0.8 +/- 0.1i

i need to find the equation, so i set the poles.

(x - (0.8 + 0.1i))(x - (0.8 - 0.1i)) = 0

multipling out (with the box method) gives:

x^2 - 0.8x - 0.1ix - 0.8x + 0.64 + 0.08i + 0.01ix - 0.08i - 0.01i^2 = 0

which = x^2 - 1.6z + 0.63 = 0

so i was like, "great ive got my equation, ill just check my results with the quadratic equation"

however, when i put the values from my equation into the quadratic equation, the value of b^2 - 4ac was positive, which would imply 2 real roots to the equation. This is obviously impossible, as the equation was determined from a complex conjugate pair!?!?!


Anyone have any suggestions on where ive gone wrong?

 

[Dr.Peterson]

Check the signs at every step of your work! Your expansion is wrong, and the discriminant for the equation you have is wrong (it isn't positive).

 

[JeffM]

When you know your method is correct and your answer is wrong, go back and look for careless mistakes. Try to avoid skipping steps.

="tjackson8684, post: 481508, member: 77866"]
we're given the complex roots are 0.8 +/- 0.1i

i need to find the equation, so i set the poles.

(x - (0.8 + 0.1i))(x - (0.8 - 0.1i)) = 0

[MATH]\implies (x - 0.8 - 0.1i)(x - 0.8 + 0.1i) = 0[/MATH]
[MATH]\implies x(x - 0.8 + 0.1i) - 0.8(x - 0.8 + 0.1i) - 0.1i(x - 0.8 + 0.1i) = 0[/MATH]
[MATH]\implies x^2 - 0.8x + 0.1ix - 0.8x + 0.64 - 0.08i - 0.1ix + 0.08i - 0.01i^2 = 0[/MATH]
[MATH]\implies x^2 - (0.8 + 0.8)x + (0.1ix - 0.1ix) + (0.08i - 0.08i) + 0.64 - 0.01(-\ 1) = 0 [/MATH]
[MATH]x^2 - 1.6x + 0.65 = 0.[/MATH]
which is NOT

x^2 - 1.6z + 0.63 = 0

so i was like, "great ive got my equation, ill just check my results with the quadratic equation"

however, when i put the values from my equation into the quadratic equation, the value of b^2 - 4ac was positive, which would imply 2 real roots to the equation. This is obviously impossible, as the equation was determined from a complex conjugate pair!?!?!

Always good to check. So lets do so with the quadratic formula (not equation).

[MATH]x = \dfrac{-\ (-\ 1.6) \pm \sqrt{(-\ 1.6)^2 - 4(1)(0.65)}}{2 * 1} = \dfrac{1.6 \pm \sqrt{2.56 - 2.6}}{2} \implies[/MATH]
[MATH]x = \dfrac{1.6 \pm \sqrt{-\ 0.04}}{2} = \dfrac{1.6 \pm \sqrt{(-\ 1)(4)(0.01)}}{2} = \dfrac{1.6 \pm 2 * 0.1 * i}{2} = 0.8 \pm 0.1i.[/MATH]
Just one tiny sign error.

 

[Steven G]

To compute (x - (0.8 + 0.1i))(x - (0.8 - 0.1i)) you should use the fact that you have conjugates (not complex conjugate roots!)
Look at (x - (0.8 + 0.1i))(x - (0.8 - 0.1i)) as ((x-.8)-.1i)((x-.8)+.1i)= (x-.8)²-(.1i)² = (x²-1.6x+.64)-(-.01) = x²-1.6x+.65

 

[tjackson8684]

OMG you guys this has literally been keeping me up at night thanks all so much. Such a frustrating mistake, and ive made it about 10 times in a row!

 

[HallsofIvy]

anyone help me out?

we're given the complex roots are 0.8 +/- 0.1i

i need to find the equation, so i set the poles.

(x - (0.8 + 0.1i))(x - (0.8 - 0.1i)) = 0

You might find it easier to do this as
((x- 0.8)+ 0.1i)((x- 0.8)- 0.1i) since then it is of the form
(a+ b)(a- b)= a^2- b^2
(x- 0.8)^2- (0.1i)^2= x^2- 0.16x+ 0.64- 0.01
= x^2- 0.16x+ 0.63.

multipling out (with the box method) gives:

x^2 - 0.8x - 0.1ix - 0.8x + 0.64 + 0.08i + 0.01ix - 0.08i - 0.01i^2 = 0

which = x^2 - 1.6z + 0.63 = 0

so i was like, "great ive got my equation, ill just check my results with the quadratic equation"

however, when i put the values from my equation into the quadratic equation, the value of b^2 - 4ac was positive, which would imply 2 real roots to the equation. This is obviously impossible, as the equation was determined from a complex conjugate pair!?!?!


Anyone have any suggestions on where ive gone wrong?

 

[Steven G]

You might find it easier to do this as
((x- 0.8)+ 0.1i)((x- 0.8)- 0.1i) since then it is of the form
(a+ b)(a- b)= a^2- b^2
(x- 0.8)^2- (0.1i)^2= x^2- 0.16x+ 0.64- 0.01
= x^2- 0.16x+ 0.63.

Please see post #4
Also your constant is wrong.
Why do we do arithmetic in public? We never learn, do we?