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Find $m$ which is a real number so that this equation has a real root.

$2z^2-(3+8i)z-(m+4i)=0$

I've tried $b^2-4ac=0 $ but I can only seem to get complex $m$ values, so either I'm missing a key point or just failing with my calculations.

Any help/ideas would really be appreciated.

Thanks :)

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  • $\begingroup$ Welcome to Math.SE! Could you include your trial using $b^2-4ac=0$, so that it might be easier for other people to help you? $\endgroup$
    – Hrodelbert
    Commented May 11, 2015 at 7:21

1 Answer 1

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HINT:

If $z$ has to be real and as $m$ is also real,

equating the real & the imaginary parts

$$2z^2-3z-m=0$$

and $$8z+4=0$$

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  • $\begingroup$ I did try this earlier and got m = 9/8 which seemed reasonable but the question was worth 20 marks so I thought there might be something more that I missed. But I guess not, thank you :) $\endgroup$
    – Evan
    Commented May 11, 2015 at 8:03
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    $\begingroup$ @ N.K. If you got $m=9/8$, you might want to recalculate. $\endgroup$
    – Hrodelbert
    Commented May 11, 2015 at 9:11

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