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# CPhill, Gavin, please help!

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

May 6, 2018

### 3+0 Answers

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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

May 6, 2018
edited by GYanggg  May 7, 2018
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Amazing! Beautiful, Gavin!

ant101  May 6, 2018
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Thanks, I'n just glad I can help.

GYanggg  May 6, 2018