Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com