Menu Close

For-the-equation-3x-2-px-3-0-find-the-value-s-of-p-if-one-root-is-i-square-of-the-other-ii-fourth-power-of-the-other-

Tinku Tara 0 Comments
Pin




Question Number 20670 by Tinkutara last updated on 31/Aug/17
For the equation 3x^2 + px + 3 = 0, find the value(s) of p if one root is (i) square of the other (ii) fourth power of the other.
$${For}\:{the}\:{equation}\:\mathrm{3}{x}^{\mathrm{2}} \:+\:{px}\:+\:\mathrm{3}\:=\:\mathrm{0}, \\ $$$${find}\:{the}\:{value}\left({s}\right)\:{of}\:{p}\:{if}\:{one}\:{root}\:{is} \\ $$$$\left({i}\right)\:{square}\:{of}\:{the}\:{other} \\ $$$$\left({ii}\right)\:{fourth}\:{power}\:{of}\:{the}\:{other}. \\ $$
Answered by dioph last updated on 31/Aug/17
(i) x_1 = x_2 ^2 1 = x_1 x_2 = x_2 ^3 ∴ x_2 = 1^(1/3) , x_1 = 1^(2/3) p = −3(x_1 +x_2 )=−3×1^(1/3) −3×1^(2/3) values of p: let ω_3 = e^(i2π/3) = (−0.5+i((√3)/2)) { ((−3×1−3×1 = −6)),((−3×ω_3 −3×ω_3 ^2 = 3)) :} (ii) x_1 = x_2 ^4 1 = x_1 x_2 = x_2 ^5 ∴ x_2 = 1^(1/5) , x_1 = 1^(4/5) p = −3(x_1 +x_2 )=−3×1^(1/5) −3×1^(4/5) values of p: let ω_5 = e^(i2π/5) = (cos ((2π)/5)+isin ((2π)/5)) if cos ((2π)/5) = (((√5)−1)/4) (from regular penthagon): { ((−3×1−3×1 = −6)),((−3×ω_5 −3×ω_5 ^4 = −6 cos ((2π)/5) = −((3((√5)−1))/2))),((−3×ω_5 ^2 −3ω_5 ^3 = 6 cos (π/5) = 3 + ((3((√5)−1))/2) )) :}
$$\left({i}\right)\:{x}_{\mathrm{1}} \:=\:{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{1}\:=\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} \:=\:{x}_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$\therefore\:{x}_{\mathrm{2}} \:=\:\mathrm{1}^{\mathrm{1}/\mathrm{3}} ,\:{x}_{\mathrm{1}} \:=\:\mathrm{1}^{\mathrm{2}/\mathrm{3}} \\ $$$${p}\:=\:−\mathrm{3}\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)=−\mathrm{3}×\mathrm{1}^{\mathrm{1}/\mathrm{3}} −\mathrm{3}×\mathrm{1}^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{values}\:\mathrm{of}\:{p}: \\ $$$$\mathrm{let}\:\omega_{\mathrm{3}} \:=\:{e}^{{i}\mathrm{2}\pi/\mathrm{3}} \:=\:\left(−\mathrm{0}.\mathrm{5}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\begin{cases}{−\mathrm{3}×\mathrm{1}−\mathrm{3}×\mathrm{1}\:=\:−\mathrm{6}}\\{−\mathrm{3}×\omega_{\mathrm{3}} −\mathrm{3}×\omega_{\mathrm{3}} ^{\mathrm{2}} \:=\:\mathrm{3}}\end{cases} \\ $$$$\left({ii}\right)\:{x}_{\mathrm{1}} \:=\:{x}_{\mathrm{2}} ^{\mathrm{4}} \\ $$$$\mathrm{1}\:=\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} \:=\:{x}_{\mathrm{2}} ^{\mathrm{5}} \\ $$$$\therefore\:{x}_{\mathrm{2}} \:=\:\mathrm{1}^{\mathrm{1}/\mathrm{5}} ,\:{x}_{\mathrm{1}} \:=\:\mathrm{1}^{\mathrm{4}/\mathrm{5}} \\ $$$${p}\:=\:−\mathrm{3}\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)=−\mathrm{3}×\mathrm{1}^{\mathrm{1}/\mathrm{5}} −\mathrm{3}×\mathrm{1}^{\mathrm{4}/\mathrm{5}} \\ $$$$\mathrm{values}\:\mathrm{of}\:{p}: \\ $$$$\mathrm{let}\:\omega_{\mathrm{5}} \:=\:{e}^{{i}\mathrm{2}\pi/\mathrm{5}} \:=\:\left(\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}+{i}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{if}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:=\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\:\left(\mathrm{from}\:\mathrm{regular}\:\mathrm{penthagon}\right):\: \\ $$$$\begin{cases}{−\mathrm{3}×\mathrm{1}−\mathrm{3}×\mathrm{1}\:=\:−\mathrm{6}}\\{−\mathrm{3}×\omega_{\mathrm{5}} −\mathrm{3}×\omega_{\mathrm{5}} ^{\mathrm{4}} \:=\:−\mathrm{6}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:=\:−\frac{\mathrm{3}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}}}\\{−\mathrm{3}×\omega_{\mathrm{5}} ^{\mathrm{2}} −\mathrm{3}\omega_{\mathrm{5}} ^{\mathrm{3}} \:=\:\mathrm{6}\:\mathrm{cos}\:\frac{\pi}{\mathrm{5}}\:=\:\mathrm{3}\:+\:\frac{\mathrm{3}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}}\:}\end{cases} \\ $$
Commented by Tinkutara last updated on 31/Aug/17
Thanks for (i) part. But in (ii) part, answer is 0 and ((3((√5) − 1))/2). Can you get it?
$$\mathrm{Thanks}\:\mathrm{for}\:\left({i}\right)\:\mathrm{part}. \\ $$$$\mathrm{But}\:\mathrm{in}\:\left({ii}\right)\:\mathrm{part},\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0}\:\mathrm{and} \\ $$$$\frac{\mathrm{3}\left(\sqrt{\mathrm{5}}\:−\:\mathrm{1}\right)}{\mathrm{2}}.\:\mathrm{Can}\:\mathrm{you}\:\mathrm{get}\:\mathrm{it}? \\ $$
Commented by dioph last updated on 31/Aug/17
I added the substitution cos ((2π)/5) = (((√5)−1)/4) but I cannot see how to get these answers If p=0 then roots are i and −i, which are not related by a fourth power
$$\mathrm{I}\:\mathrm{added}\:\mathrm{the}\:\mathrm{substitution} \\ $$$$\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:=\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{see}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{these}\:\mathrm{answers} \\ $$$$\mathrm{If}\:{p}=\mathrm{0}\:\mathrm{then}\:\mathrm{roots}\:\mathrm{are}\:{i}\:\mathrm{and}\:−{i}, \\ $$$$\mathrm{which}\:\mathrm{are}\:\mathrm{not}\:\mathrm{related}\:\mathrm{by}\:\mathrm{a}\: \\ $$$$\mathrm{fourth}\:\mathrm{power} \\ $$
Commented by Tinkutara last updated on 31/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *