First slide

Theory of equations

Question

If $\frac{1}{3 - 4 i}$ is a root of $a x^{2} + b x + c = 0, (a, b$
$c \in R, a \neq 0)$ , then

Moderate

A

$b + 6 c = 0$
B

$b = 6 c$
C

$a + 25 c = 0$
D

$b^{2} = 6 c$

Solution

TIP: As $a, b, c \in R$ the other root must $\frac{1}{3 + 4 i}$
Thus, $\frac{1}{3 - 4 i} + \frac{1}{3 + 4 i} = - \frac{b}{a},$
and $\frac{1}{3 - 4 i} \cdot \frac{1}{3 + 4 i} = \frac{c}{a}$
$\Rightarrow \frac{6}{25} = - \frac{b}{a} and \frac{1}{25} = \frac{c}{a}$
Now, $6 (\frac{c}{a}) = - \frac{b}{a} \Rightarrow b + 6 c = 0$
Also, $a - 25 c = 0$ .

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