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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-pu

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Question Number 51843 by prakash jain last updated on 31/Dec/18
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.
$${If}\:{ax}^{\mathrm{2}} +{bx}+{c}+{i}=\mathrm{0}\:\mathrm{has}\:\mathrm{purely} \\ $$$$\mathrm{imaginary}\:\mathrm{roots}\:\mathrm{where}\: \\ $$$${a},{b},{c}\:{are}\:{non}−{zero}\:{real}. \\ $$$${answer}\:{given}:\:{a}={b}^{\mathrm{2}} {c} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{question}\:\mathrm{is}\:\mathrm{wrong} \\ $$$$\mathrm{since}\:\mathrm{if}\:{z}_{\mathrm{1}} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{roots}\:\mathrm{than} \\ $$$${z}_{\mathrm{1}} +{z}_{\mathrm{2}} =−\frac{{b}}{{a}} \\ $$$${purely}\:{imaginary}={purely}\:{real} \\ $$$${not}\:{possible} \\ $$$$\mathrm{Can}\:\mathrm{some}\:\mathrm{point}\:\mathrm{a}\:\mathrm{mistake}. \\ $$
Commented by Rasheed.Sindhi last updated on 31/Dec/18
Sir Prakash Jain! Very Happy to see you after long time on this platform! I remember the period led by you and Yozzi on this platform! The period which was full of conceptual discussions and topic based long sessions!
$$\mathcal{S}{ir}\:\mathcal{P}{rakash}\:\mathcal{J}{ain}! \\ $$$$\mathcal{V}{ery}\:\mathcal{H}{appy}\:{to}\:{see}\:{you}\:{after} \\ $$$${long}\:{time}\:{on}\:{this}\:{platform}! \\ $$$$\mathcal{I}\:{remember}\:{the}\:{period}\:\boldsymbol{{led}}\:{by}\:{you} \\ $$$${and}\:{Yozzi}\:{on}\:{this}\:{platform}!\:\mathcal{T}{he} \\ $$$${period}\:\:{which}\:{was}\:{full}\:{of}\:{conceptual} \\ $$$${discussions}\:{and}\:{topic}\:{based}\:{long}\:{sessions}! \\ $$
Commented by Rasheed.Sindhi last updated on 31/Dec/18
May I be so lucky to learn morefrom you!
$$\mathcal{M}{ay}\:\mathcal{I}\:{be}\:{so}\:{lucky}\:{to}\:{learn}\:{morefrom} \\ $$$${you}! \\ $$
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
let x = iq −aq^2 +ibq+c+i= 0 ⇒ aq^2 = c & bq+1= 0 ⇒ a = b^2 c .
$${let}\:{x}\:=\:{iq} \\ $$$$−{aq}^{\mathrm{2}} +{ibq}+{c}+{i}=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{aq}^{\mathrm{2}} =\:{c}\:\:\:\&\:\:\:{bq}+\mathrm{1}=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{a}\:=\:{b}^{\mathrm{2}} {c}\:. \\ $$
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 cx^2 +bx+c+i=0 x^2 +(x/(bc))+((c+i)/(b^2 c))=0 clearly x=−(i/b) is a root as you calculated. let other root be z. (x+(i/b))(x−z)=0 (i/b)−z=(1/(bc))⇒z=(i/b)−(1/(bc)) (1/(bc)) is real so other root is not purely imaginary.
$$\mathrm{This}\:\mathrm{is}\:\mathrm{valid}\:\mathrm{only}\:\mathrm{when}\:\mathrm{question}\:\mathrm{says} \\ $$$$\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{is}\:\mathrm{imaginary}. \\ $$$${a}={b}^{\mathrm{2}} {c} \\ $$$${b}^{\mathrm{2}} {cx}^{\mathrm{2}} +{bx}+{c}+{i}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\frac{{x}}{{bc}}+\frac{{c}+{i}}{{b}^{\mathrm{2}} {c}}=\mathrm{0} \\ $$$$\mathrm{clearly}\:{x}=−\frac{{i}}{{b}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{as}\:\mathrm{you} \\ $$$$\mathrm{calculated}. \\ $$$$\mathrm{let}\:\mathrm{other}\:\mathrm{root}\:\mathrm{be}\:{z}. \\ $$$$\left({x}+\frac{{i}}{{b}}\right)\left({x}−{z}\right)=\mathrm{0} \\ $$$$\frac{{i}}{{b}}−{z}=\frac{\mathrm{1}}{{bc}}\Rightarrow{z}=\frac{{i}}{{b}}−\frac{\mathrm{1}}{{bc}} \\ $$$$\frac{\mathrm{1}}{{bc}}\:\mathrm{is}\:\mathrm{real}\:\mathrm{so}\:\mathrm{other}\:\mathrm{root}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{purely}\:\mathrm{imaginary}. \\ $$

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