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The question is as follows;

Suppose $2+7i$ is a solution of $2z^2+Az+B=0$, where $A, B \in \mathbb{R}$ . Find $A$ and $B$.

My understanding is that this equation holds:

$$2(2+7i)^2 + A(2+7i) + B = 0$$

which will eventually lead to:

$$-90 + 2A + B + i(56+7A) = 0$$

I would like to check if my approach is correct, and if so, what should I do next to derive $A$ & $B$.

X-men
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    If $A,B\in\mathbb R$, then $\overline z=2-7i$. Finally, you can apply Vieta's formula. – lone student Sep 04 '21 at 07:08
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    you can realize that $0 = 0 + 0i$, equating the real and imaginary parts you get a system of equations in the unknowns $A$ and $B$. – Colver Sep 04 '21 at 07:11
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    @colver, solved it ! I was so focused on solving the real equation that i didn't realise i could solve for Imaginary part thus leading to the real part. Thank you! – X-men Sep 04 '21 at 07:49
  • An approach by Lalit Tolani: since $A,B$ are real, the roots must come in conjugate pairs, or that $-\frac{A}{2} = (2 + 7i) + (2 - 7i), \frac{B}{2} = (2 + 7i)(2 - 7i)$ from Vieta's formulas. – Toby Mak Sep 04 '21 at 08:43

3 Answers3

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$−90+2A+B+i(56+7A)=0$

Since $0 + 0i = 0$,

we can equate the following.

$i(56 + 7A) = 0$

$7A = -56$, thus $A = -8$

Sub A into the Real Part of the equation to get B.

Real part of the equation $-90+2(A)+B=0$

You should derive 106 for B.

Huge thanks to Colver for the heads up!

X-men
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Note that the question gives hint :

If $A$ and $B$ are real, then complex roots occur in conjugate pairs, therefore If one root is $2+7i$ other will be $2-7i$.

Now Sum of roots , $-\frac{A}{2}=4\implies A=-8$

Product of roots , $\frac{B}{2}=4+49\implies B=106 $

Lalit Tolani
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A polynomial with real coefficients will have non-real roots in conjugate pairs. Since it has $2+7i$ as a root, it will also have $2-7i$ as a root. Being a degree $2$ polynomial, these are the only roots. The leading coefficient of $2z^2+Az+B=0$ is $2$, so our polynomial is: $$2(z- (2-7i))(z -(2+7i)) = 0$$ Expanding, the coefficients we obtain: $$2 z^2 - 2(2-7i + 2+7i)z + 2(2-7i)(2+7i) = 0\\ 2z^2 - 8z + 106 = 0$$ We get: $A=-8$, $B = 106$.

user3257842
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