complex quadratic equation (1 Viewer)

elliottjohnson · Feb 15, 2010

10 (c) p30 arnold

1-2i is one root of x^2-(3+i)x+k = 0.

Find k and the other root of the equation.

Why is the other root NOT the conjugate of 1-2i ???
under what conditions is this rule valid??

thanks for any clearing up.

xV1P3R · Feb 15, 2010

Rule is valid when all the coefficients are real.
$$1-2i is one root of x^2-(3+i)x+k = 0$ \\$You know one of the roots $ \alpha =1-2i \\\sum \alpha =\alpha +\beta =\frac{-b}{a} \\1-2i+\beta =\frac{3+i}{1} \\\beta =2+3i \\\sum \alpha \beta =\frac{c}{a} \\(1-2i)(2+3i)=\frac{k}{1} \\2+3i-4i+6=k \\k=8-i$

elliottjohnson · Feb 15, 2010

fantastic thanks!!!

untouchablecuz · Feb 15, 2010

alternatively:

$\\P(x)=x^2-(3+i)x+k\\$If $1-2i $ is a root of $P(x) $ then $P(1-2i)=0,\\(1-2i)^2-(3+i)(1-2i)+k \\=1-4i-4-3+6i-i-2+k\\=-10+i+k\\=0\\ \fbox {$k=10-i }$

xV1P3R · Feb 15, 2010

untouchablecuz said:
alternatively:

$\\P(x)=x^2-(3+i)x+k\\$If $1-2i $ is a root of $P(x) $ then $P(1-2i)=0,\\(1-2i)^2-(3+i)(1-2i)+k \\=1-4i-4-3+6i-i-2+k\\=-10+i+k\\=0\\ \fbox {$k=10-i }$

Yeah reckon this one's better. Mistake in adding reals in 4th from bottom.

complex quadratic equation (1 Viewer)

elliottjohnson

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