+0  
 
+1
80
3
avatar+493 

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
 #1
avatar+867 
+9

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

GYanggg  May 6, 2018
edited by GYanggg  May 7, 2018
 #2
avatar+493 
+5

Amazing! Beautiful, Gavin! 

ant101  May 6, 2018
 #3
avatar+867 
+8

Thanks, I'n just glad I can help. 

GYanggg  May 6, 2018

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