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Question Number 165651 by MathsFan last updated on 06/Feb/22
If the equation (2x−1)−p(x^2 +2)=0, where p is constant, has imaginary roots, deduce that 2p^2 +p−1≥0.
Iftheequation(2x−1)−p(x2+2)=0,wherepisconstant,hasimaginaryroots,deducethat2p2+p−1⩾0.
Commented by mr W last updated on 06/Feb/22
px^2 −2x+2p+1=0 x=((2±(√(2^2 −4p(2p+1))))/(2p))=((1±(√(1−p−2p^2 )))/p) if 1−p−2p^2 ≥0 ⇒real roots if 1−p−2p^2 <0 ⇒complex roots: x=((1±i(√(2p^2 +p−1)))/p) since (1/p)≠0, the equation can never have imaginary roots! therefore: question is wrong! 1) when 2p^2 +p−1=0 it has real roots. 2) when 2p^2 +p−1>0 it has comlex roots.
px2−2x+2p+1=0x=2±22−4p(2p+1)2p=1±1−p−2p2pif1−p−2p2⩾0⇒realrootsif1−p−2p2<0⇒complexroots:x=1±i2p2+p−1psince1p≠0,theequationcanneverhaveimaginaryroots!therefore:questioniswrong!1)when2p2+p−1=0ithasrealroots.2)when2p2+p−1>0ithascomlexroots.
Commented by peter frank last updated on 06/Feb/22
thank you
thankyou
Commented by MathsFan last updated on 06/Feb/22
thanks
thanks
Commented by chrisbridge last updated on 06/Feb/22
i think you do not need (1/p) to be equal to 0 to have imaginary roots. You only need the discrimant to be = 0. From your work above 2p^2 +p−1>0 is true for the first equation to have imaginary roots
ithinkyoudonotneed1ptobeequalto0tohaveimaginaryroots.Youonlyneedthediscrimanttobe=0.Fromyourworkabove2p2+p−1>0istrueforthefirstequationtohaveimaginaryroots
Commented by mr W last updated on 06/Feb/22
i don′t need, but the question needs! the question requests imaginary roots. bi is an imaginary number. a+bi is a complex number, not an imaginary number, if a≠0. since (1/p) can′t be zero, so the equation can never have imaginary roots!
idon′tneed,butthequestionneeds!thequestionrequestsimaginaryroots.biisanimaginarynumber.a+biisacomplexnumber,notanimaginarynumber,ifa≠0.since1pcan′tbezero,sotheequationcanneverhaveimaginaryroots!
Answered by chrisbridge last updated on 06/Feb/22
b^2 −4ac<0 for imaginary roots 2x − 1 −px^2 −2p = 0 px^2 −2x+(1+2p)=0 (−2)^2 −4p(1+2p)<0 4−4p−8p^2 <0 4−4p−8p^2 <0 dividing through by 4 1−p−2p^2 <0 on multiplying through by (−1) −1+p+2p^2 >0 rearraging 2p^2 +p−1>0 required solution
b2−4ac<0forimaginaryroots2x−1−px2−2p=0px2−2x+(1+2p)=0(−2)2−4p(1+2p)<04−4p−8p2<04−4p−8p2<0dividingthroughby41−p−2p2<0onmultiplyingthroughby(−1)−1+p+2p2>0rearraging2p2+p−1>0requiredsolution
Commented by mr W last updated on 07/Feb/22
you seem to think that imaginary numbers and complex numbers are the same thing.
youseemtothinkthatimaginarynumbersandcomplexnumbersarethesamething.

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