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InterviewSolution
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Consider a quadratic equaiton `az^(2) + bz + c=0`, where a,b,c are complex number. If equaiton has two purely imaginary roots, then which of the following is not ture.A. `abarb` is purely imaginaryB. `b barc` is purely imaginaryC. `cbara` is purely realD. none of these |
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Answer» Correct Answer - D Let `z_(1)` (purely imginary ) be a root of the given equation Then, `z_(1) = - barz_(1)` and `underline(az_(1)^(2) + bz_(1) + c)=0" "(1)` `rArr az_(1)^(2) + bz_(1) + c = 0` `rArr bara barz_(1)^(2) + barb barz_(1) + c = 0` `rArr bar z bar z_(1)^(2) + bar b bar z_(1) + barc = 0` `rArr bar a bar z_(1)^(2) - bar b barz_(1) + barc = 0" "(as barz_(1) = - z_(1))" "(2)` Now Eqs. (1) and (2) must have one common root. `therefore ( cbara-abarc)^(2) = (barbc+ cbarb) (-abarb - barab)` Let `z_(1)` and `z_(2)` be two purely imaginary roots. Then, `barz_(1) = -z_(1), barz_(2) = - z_(2)` Now , `underline(abarz^(2) + bz + c) = 0" "(3)` or `az^(2) + bz + c=bar0` or `bara barz_(20 + barb barz + barc =0` or `bara z^(2) - barbz + barc = 0" "(4)` Equations (3) and (4) must be identical as their roots are same. ` therefore (a)/(bara) = -(b)/(barb) =(c)/(barc)` `rArr abarc = barac, + barab = 0` and `b barc +barbc=0` . Hence, `barac` is purely real and `abarb` and `bbarc` are purely imaginary . let `z_(1)` (purely real ) be a root of the given equation . Then , `z_(1) = barz_(1)` ltbr gt and ``underline(az_(1)^(2) + bz_(1) + c)= bar0" "(5)` or `az_(1)^(2) + bz_(1) + c=0` or `baraz_(1)^(2) + bz_(1) + c = bar0` or `baraz_(1)^(2) + barb z_(1) + c= 0" "(6)` Now(5) and (6) must have one common root. Hence, `(cbara - abarc)^(2) = (b barc - cbarb)(abarb-barab)` |
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