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The roots of the quadratic equation $z^2 + bz + c = 0$ are $-7 + 2i$ and $-7 - 2i$. What is $b+c$?

Sagot :

dasorin7

Answer:

67

Step-by-step explanation: Given the quadratic equation $z^2 + bz + c = 0$, Vieta's formulas tell us the sum of the roots is $-b$, and the product of the roots is $c$. Thus,

\[-b = (-7 + 2i) + (-7 - 2i) = -14,\]so $b = 14.$

Also,

\[c = (-7 + 2i)(-7 - 2i) = (-7)^2 - (2i)^2 = 49 + 4 = 53.\]Therefore, we have $b+c = \boxed{67}$.

There are many other solutions to this problem. You might have started with the factored form $(z - (-7 + 2i))(z - (-7 - 2i)),$ or even thought about the quadratic formula.

This is the aops answer :)

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