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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.

$${If}\:{the}\:{equation}\:\left(\mathrm{2}{x}−\mathrm{1}\right)−{p}\left({x}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0}, \\ $$$${where}\:{p}\:{is}\:{constant},\:{has}\:{imaginary} \\ $$$${roots},\:{deduce}\:{that}\:\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}\geqslant\mathrm{0}. \\ $$

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.

$${px}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}{p}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}\pm\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{4}{p}\left(\mathrm{2}{p}+\mathrm{1}\right)}}{\mathrm{2}{p}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−{p}−\mathrm{2}{p}^{\mathrm{2}} }}{{p}} \\ $$$${if}\:\mathrm{1}−{p}−\mathrm{2}{p}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow{real}\:{roots} \\ $$$${if}\:\mathrm{1}−{p}−\mathrm{2}{p}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow{complex}\:{roots}: \\ $$$$\:\:\:\:{x}=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}}}{{p}} \\ $$$$\:\:\:\:{since}\:\frac{\mathrm{1}}{{p}}\neq\mathrm{0},\:{the}\:{equation}\:{can}\:{never} \\ $$$$\:\:\:\:{have}\:{imaginary}\:{roots}! \\ $$$$ \\ $$$${therefore}:\:{question}\:{is}\:{wrong}! \\ $$$$\left.\mathrm{1}\right)\:{when}\:\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}=\mathrm{0}\:{it}\:{has}\:{real}\:{roots}. \\ $$$$\left.\mathrm{2}\right)\:{when}\:\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}>\mathrm{0}\:{it}\:{has}\:{comlex}\:{roots}. \\ $$

Commented by peter frank last updated on 06/Feb/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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

$${i}\:{think}\:{you}\:{do}\:{not}\:{need}\:\frac{\mathrm{1}}{{p}}\:{to}\:{be}\:{equal}\:{to}\:\mathrm{0}\:{to}\:{have}\:{imaginary}\:{roots}.\:{You}\:{only}\:{need}\:{the}\:{discrimant}\:{to}\:{be}\:=\:\mathrm{0}.\:{From}\:{your}\:{work}\:{above}\:\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}>\mathrm{0}\:{is}\:{true}\:{for}\:{the}\:{first}\:{equation}\:{to}\:{have}\:{imaginary}\:{roots} \\ $$

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!

$${i}\:{don}'{t}\:{need},\:{but}\:{the}\:{question}\:{needs}! \\ $$$${the}\:{question}\:{requests}\:{imaginary}\:{roots}. \\ $$$${bi}\:{is}\:{an}\:{imaginary}\:{number}. \\ $$$${a}+{bi}\:{is}\:{a}\:{complex}\:{number},\:{not}\:{an} \\ $$$${imaginary}\:{number},\:{if}\:{a}\neq\mathrm{0}. \\ $$$${since}\:\frac{\mathrm{1}}{{p}}\:{can}'{t}\:{be}\:{zero},\:{so}\:{the}\:{equation} \\ $$$${can}\:{never}\:{have}\:{imaginary}\:{roots}! \\ $$

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

$${b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0}\:{for}\:{imaginary}\:{roots} \\ $$$$\mathrm{2}{x}\:−\:\mathrm{1}\:−{px}^{\mathrm{2}} \:−\mathrm{2}{p}\:=\:\mathrm{0} \\ $$$${px}^{\mathrm{2}} −\mathrm{2}{x}+\left(\mathrm{1}+\mathrm{2}{p}\right)=\mathrm{0} \\ $$$$\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}{p}\left(\mathrm{1}+\mathrm{2}{p}\right)<\mathrm{0} \\ $$$$\mathrm{4}−\mathrm{4}{p}−\mathrm{8}{p}^{\mathrm{2}} <\mathrm{0} \\ $$$$\mathrm{4}−\mathrm{4}{p}−\mathrm{8}{p}^{\mathrm{2}} <\mathrm{0} \\ $$$${dividing}\:{through}\:{by}\:\mathrm{4} \\ $$$$\mathrm{1}−{p}−\mathrm{2}{p}^{\mathrm{2}} <\mathrm{0} \\ $$$${on}\:{multiplying}\:{through}\:{by}\:\left(−\mathrm{1}\right) \\ $$$$−\mathrm{1}+{p}+\mathrm{2}{p}^{\mathrm{2}} >\mathrm{0} \\ $$$${rearraging} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}>\mathrm{0} \\ $$$${required}\:{solution} \\ $$$$ \\ $$$$ \\ $$

Commented by mr W last updated on 07/Feb/22

you seem to think that imaginary numbers and complex numbers are the same thing.

$${you}\:{seem}\:{to}\:{think}\:\:{that}\:{imaginary}\: \\ $$$${numbers}\:{and}\:{complex}\:{numbers}\:{are}\: \\ $$$${the}\:{same}\:{thing}. \\ $$

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