If ax^2 +bx+c+i=0 has purely imaginary roots where a,b,c are non−zero real. answer given: a=b^2 c I think question is wrong since if z_1 and z_2 are roots than z_1 +z_2 =−(b/a) purely imaginary=purely real not possible Can some point a mistake.


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Question Number 51843 by prakash jain last updated on 31/Dec/18

$I f {a x}^{2} + b x + c + i = 0 has purely$ $imaginary roots where$ $a, b, c a r e n o n - z e r o r e a l .$ $a n s w e r g i v e n : a = b^{2} c$ $I think question is wrong$ $since if z_{1} and z_{2} are roots than$ $z_{1} + z_{2} = - \frac{b}{a}$ $p u r e l y i m a g i n a r y = p u r e l y r e a l$ $n o t p o s s i b l e$ $Can some point a mistake .$
Commented by Rasheed.Sindhi last updated on 31/Dec/18

$S i r P r a k a s h J a i n!$ $V e r y H a p p y t o s e e y o u a f t e r$ $l o n g t i m e o n t h i s p l a t f o r m!$ $I r e m e m b e r t h e p e r i o d l e d b y y o u$ $a n d Y o z z i o n t h i s p l a t f o r m! T h e$ $p e r i o d w h i c h w a s f u l l o f c o n c e p t u a l$ $d i s c u s s i o n s a n d t o p i c b a s e d l o n g s e s s i o n s!$
Commented by Rasheed.Sindhi last updated on 31/Dec/18

$M a y I b e s o l u c k y t o l e a r n m o r e f r o m$ $y o u!$
Commented by prakash jain last updated on 03/Jan/19
Hi Rasheed, good to see you are still around. I am learning from you and everyone else in the forum.
	Answered by ajfour last updated on 31/Dec/18

	$l e t x = i q$ $- {a q}^{2} + i b q + c + i = 0$ $\Rightarrow {a q}^{2} = c & b q + 1 = 0$ $\Rightarrow a = b^{2} c .$
	Commented by prakash jain last updated on 31/Dec/18

	$This is valid only when question says$ $one of the roots is imaginary .$ $a = b^{2} c$ $b^{2} {c x}^{2} + b x + c + i = 0$ $x^{2} + \frac{x}{b c} + \frac{c + i}{b^{2} c} = 0$ $clearly x = - \frac{i}{b} is a root as you$ $calculated .$ $let other root be z .$ $(x + \frac{i}{b}) (x - z) = 0$ $\frac{i}{b} - z = \frac{1}{b c} \Rightarrow z = \frac{i}{b} - \frac{1}{b c}$ $\frac{1}{b c} is real so other root is not$ $purely imaginary .$
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