Complex Conjugates Question (1 Viewer)

Ambility · Oct 13, 2015

Question 5a and 5b from 2.1 in the Cambridge 4 Unit Maths textbook:

$$5a. $a\alpha^2+b\alpha+c=0$, where a,b,c are real and $\alpha$ is complex. Show that $a\bar{\alpha}^2+b\bar{\alpha}+c=0.$

I remember hearing that the roots to a quadratic are complex conjugates of each other, but I don't know how I'd prove this. Could someone point me in the right direction?

Carrotsticks · Oct 13, 2015

Ambility said:
Question 5a and 5b from 2.1 in the Cambridge 4 Unit Maths textbook:

$$5a. $a\alpha^2+b\alpha+c=0$, where a,b,c are real and $\alpha$ is complex. Show that $a\bar{\alpha}^2+b\bar{\alpha}+c=0.$

I remember hearing that the roots to a quadratic are complex conjugates of each other, but I don't know how I'd prove this. Could someone point me in the right direction?

Start with the given expression and conjugate both sides. From there, use your conjugate rules to slowly 'spread' the conjugate symbol across all terms to obtain the required result.

Drsoccerball · Oct 13, 2015

Ambility said:
Question 5a and 5b from 2.1 in the Cambridge 4 Unit Maths textbook:

$$5a. $a\alpha^2+b\alpha+c=0$, where a,b,c are real and $\alpha$ is complex. Show that $a\bar{\alpha}^2+b\bar{\alpha}+c=0.$

I remember hearing that the roots to a quadratic are complex conjugates of each other, but I don't know how I'd prove this. Could someone point me in the right direction?

$$ Equation with roots $ \alpha $ can be expressed as $ 0=ax^2 +bx +c.$

$Since a,b and c are all real and $ \alpha $ is a root via conjugate root theorem conjugate is also a root$

$\therefore a\bar{\alpha}^2 +b\bar{\alpha} +c = 0$

Ambility · Oct 13, 2015

Carrotsticks said:
Start with the given expression and conjugate both sides. From there, use your conjugate rules to slowly 'spread' the conjugate symbol across all terms to obtain the required result.

Oh right, and I guess the complex conjugate of 0 is just 0?

Carrotsticks · Oct 13, 2015

Drsoccerball said:
$$ Equation with roots $ \alpha $ can be expressed as $ 0=ax^2 +bx +c.$

$Since a,b and c are all real and $ \alpha $ is a root via conjugate root theorem conjugate is also a root$

$\therefore a\bar{\alpha}^2 +b\bar{\alpha} +c = 0$

He is trying to prove the conjugate root theorem for polynomials of degree 2. Stating the theorem to prove it is a bit redundant. Your response would be more appropriate for a question styled as "Explain why ....", not "Show that ..."

Ambility said:
Oh right, and I guess the complex conjugate of 0 is just 0?

Yes.

Complex Conjugates Question (1 Viewer)

Ambility

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Carrotsticks

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Drsoccerball

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Ambility

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Carrotsticks

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