+884 The roots of the quadratic equation \(z^2 + az + b = 0\) are \(2-3i\) and \(2+3i\). What is \(a+b\)?
+981 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 \(a^2-b^2=(a-b)(a+b)\),
\((2-3i)(2+3i)=2^2-(3i)^2=4-(9\cdot(-1))=4+9=13\)
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,
Gavin
