quadratic equations with imaginary numbers

**edm** · August 7th 2009, 02:07 PM

I am doing some work with z transforms which are littered with quadratics, but I am slightly confused about a fundamental when solving a quadratics with an imaginary component.

$2s^2-2s+1 = 0$

$\frac{2\pm\sqrt{-4}} {4} = \frac{2\pm2j} {4} = \frac{1} {2} \pm \frac{1}{2}j$

thats fine. no problem. its next example which has me confused,

$s^2 + 2s + 5 = 0$

$\frac {-2\pm \sqrt{-16}} {10} = \frac {-2 \pm 4j} {10} = -\frac{1}{5} \pm \frac{2}{5}j$

but the answer should be,

$-1 \pm 2j$

so what happens to the denominator?

thanks!

**Plato** · August 7th 2009, 02:23 PM

Originally Posted by edm

its next example which has me confused,

$s^2 + 2s + 5 = 0$

$\frac {-2\pm \sqrt{-16}} {{\color{red}10}} = \frac {-2 \pm 4j} {{\color{red}10}} = -\frac{1}{5} \pm \frac{2}{5}j$

but the answer should be, so what happens to the denominator? $-1 \pm 2j$

The denominator is 2 not 10.

**Stroodle** · August 7th 2009, 02:24 PM

Remember that the denominator in the quadratic formula is $2a$ from the general quadratic equation $ax^2+bx+c=0$

**edm** · August 7th 2009, 03:02 PM

ah bloody hell thats why it made no sense, i took the s or a 5

.

thats the problem when you scribble down Ss 2s and Zs they all look the same (well in my hand writing anyway!).

many thanks.

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