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

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:

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

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

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

• If $A,B\in\mathbb R$, then $\overline z=2-7i$. Finally, you can apply Vieta's formula. Commented Sep 4, 2021 at 7:08
• you can realize that $0 = 0 + 0i$, equating the real and imaginary parts you get a system of equations in the unknowns $A$ and $B$. Commented Sep 4, 2021 at 7:11
• @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! Commented Sep 4, 2021 at 7:49
• 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. Commented Sep 4, 2021 at 8:43

$$−90+2A+B+i(56+7A)=0$$

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!

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

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: $$A=-8$$, $$B = 106$$.