The roots of the quadratic equation \(z^2 + az + b = 0\) are \(2-3i\)  and \(2+3i\). What is \(a+b\)?

ant101  May 6, 2018

Hi ant101!


We use the rule that states:


In the quadratic equation \(ax^2+bx+c=0\)


The sum of the roots is \(-\frac{b}{a}\)


The produdct of the roots is \(\frac{c}{a}\)


In your question, the sum of the two roots is \(2-3i+3i+2=4\)


The product is, \((2-3i)(2+3i)\)


Using \(a^2-b^2=(a-b)(a+b)\)




Using the law, \(-\frac{a}{1}=4 \\ \frac{b}{1}=13\)


\(a=-4 \\ b=13 \\ 13+(-4)=\boxed9\)


That is the answer you seek.


I hope this helped,


GYanggg  May 6, 2018
edited by GYanggg  May 7, 2018

Amazing! Beautiful, Gavin! 

ant101  May 6, 2018

Thanks, I'n just glad I can help. 

GYanggg  May 6, 2018

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