+0  
 
0
32
1
avatar

The roots of the quadratic equation z^2 + az + b = 0 are -7 + 3i and -7 - 3i.  What is a + b?

 Jul 2, 2021
 #1
avatar+234 
0

The quadratic formula is, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\). First let's focus on the -7. To make -b/2 = -7, b would have to be b = 14 to satisfy. So already have,

\(z^2 + 14z + b = 0\)

Now that we know b, we need to have \(\frac{\sqrt{196-4c}}{2} = 3i\). Square both sides and further solve, 

\(\frac{196 - 4c}{4} = -9\)

\(196 - 4c = -36\)

\(4c = 232\)

\(c = 58\)

 

So completing the equation we have, \(z^2 + 14z + 58 = 0\), and we can check using the quadratic formula to see that the roots are \(-7 \pm 3i\). So,\( a + b = 14 + 58 =\) 72

 Jul 3, 2021

3 Online Users