Menu Close

x-3-9xy-2-28-x-2-y-y-3-15-solve-for-x-and-y-x-y-R-

Tinku Tara 0 Comments
Pin




Question Number 209761 by pablo1234523 last updated on 20/Jul/24
x^3 −9xy^2 =28 x^2 y−y^3 =15 solve for x and y. x,y∈R
$${x}^{\mathrm{3}} −\mathrm{9}{xy}^{\mathrm{2}} =\mathrm{28} \\ $$$${x}^{\mathrm{2}} {y}−{y}^{\mathrm{3}} =\mathrm{15} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x}\:\mathrm{and}\:{y}. \\ $$$${x},{y}\in\mathbb{R} \\ $$
Answered by mr W last updated on 20/Jul/24
let y=kx x^3 (1−9k^2 )=28 ...(i) x^3 (k−k^3 )=15 ...(ii) (i)/(ii): ((1−9k^2 )/(k−k^3 ))=((28)/(15)) 28k^3 −135k^2 −28k+15=0 (k−5)(4k−1)(7k+3)=0 ⇒k=−(3/7), (1/4), 5 ⇒x=(((28)/(1−9k^2 )))^(1/3) =−(7/2), 4, −(1/2) ⇒y=(3/2), 1, −(5/2)
$${let}\:{y}={kx} \\ $$$${x}^{\mathrm{3}} \left(\mathrm{1}−\mathrm{9}{k}^{\mathrm{2}} \right)=\mathrm{28}\:\:\:…\left({i}\right) \\ $$$${x}^{\mathrm{3}} \left({k}−{k}^{\mathrm{3}} \right)=\mathrm{15}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\frac{\mathrm{1}−\mathrm{9}{k}^{\mathrm{2}} }{{k}−{k}^{\mathrm{3}} }=\frac{\mathrm{28}}{\mathrm{15}} \\ $$$$\mathrm{28}{k}^{\mathrm{3}} −\mathrm{135}{k}^{\mathrm{2}} −\mathrm{28}{k}+\mathrm{15}=\mathrm{0} \\ $$$$\left({k}−\mathrm{5}\right)\left(\mathrm{4}{k}−\mathrm{1}\right)\left(\mathrm{7}{k}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=−\frac{\mathrm{3}}{\mathrm{7}},\:\frac{\mathrm{1}}{\mathrm{4}},\:\mathrm{5} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{28}}{\mathrm{1}−\mathrm{9}{k}^{\mathrm{2}} }}=−\frac{\mathrm{7}}{\mathrm{2}},\:\mathrm{4},\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{3}}{\mathrm{2}},\:\mathrm{1},\:−\frac{\mathrm{5}}{\mathrm{2}} \\ $$
Commented by pablo1234523 last updated on 20/Jul/24
thanks a lot sir; how did u find the roots of that cubic equation? by inspection i think. my actual question was: − how to find ((28+45i(√3)))^(1/3) .
$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{sir}; \\ $$$$\mathrm{how}\:\mathrm{did}\:\mathrm{u}\:\mathrm{find}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{that}\:\mathrm{cubic}\:\mathrm{equation}? \\ $$$$\mathrm{by}\:\mathrm{inspection}\:\mathrm{i}\:\mathrm{think}. \\ $$$$\mathrm{my}\:\mathrm{actual}\:\mathrm{question}\:\mathrm{was}:\:− \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\sqrt[{\mathrm{3}}]{\mathrm{28}+\mathrm{45}{i}\sqrt{\mathrm{3}}}. \\ $$
Commented by mr W last updated on 20/Jul/24
i tried with the factors of 15, and found 5 is a root. the rest is then easy.
$${i}\:{tried}\:{with}\:{the}\:{factors}\:{of}\:\mathrm{15},\:{and} \\ $$$${found}\:\mathrm{5}\:{is}\:{a}\:{root}.\:{the}\:{rest}\:{is}\:{then} \\ $$$${easy}. \\ $$
Commented by mr W last updated on 20/Jul/24
28+45(√3)i=((√(19)))^3 e^(itan^(−1) ((45(√3))/(28))) ((28+45(√3)i))^(1/3) =(√(19))e^(i(((2kπ)/3)+(1/3)tan^(−1) ((45(√3))/(28)))) with k=0,1,2
$$\mathrm{28}+\mathrm{45}\sqrt{\mathrm{3}}{i}=\left(\sqrt{\mathrm{19}}\right)^{\mathrm{3}} {e}^{{i}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{45}\sqrt{\mathrm{3}}}{\mathrm{28}}} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{28}+\mathrm{45}\sqrt{\mathrm{3}}{i}}=\sqrt{\mathrm{19}}{e}^{{i}\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{45}\sqrt{\mathrm{3}}}{\mathrm{28}}\right)} \:{with}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$
Commented by pablo1234523 last updated on 20/Jul/24
thanks sir, infact ((28+45i(√3)))^(1/3) =−(7/2)+((3i(√3))/2), 4+i(√3), −(1/2)−((5i(√3))/2) is there method to find tan((1/3)tan^(−1) ((45(√3))/(28))) more specifucally, cos ((1/3)tan^(−1) ((45(√3))/(28)) )=cos ((1/3)cos^(−1) ((28)/(19(√(19)))) ) similarly for sin(..)
$$\mathrm{thanks}\:\mathrm{sir},\: \\ $$$$\mathrm{infact}\:\sqrt[{\mathrm{3}}]{\mathrm{28}+\mathrm{45i}\sqrt{\mathrm{3}}}=−\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{3}{i}\sqrt{\mathrm{3}}}{\mathrm{2}},\:\mathrm{4}+{i}\sqrt{\mathrm{3}},\:−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{5}{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{is}\:\mathrm{there}\:\mathrm{method}\:\mathrm{to}\:\mathrm{find}\:\mathrm{tan}\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{45}\sqrt{\mathrm{3}}}{\mathrm{28}}\right) \\ $$$$\mathrm{more}\:\mathrm{specifucally},\:\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{45}\sqrt{\mathrm{3}}}{\mathrm{28}}\:\right)=\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{28}}{\mathrm{19}\sqrt{\mathrm{19}}}\:\right) \\ $$$$\mathrm{similarly}\:\mathrm{for}\:\mathrm{sin}\left(..\right) \\ $$$$ \\ $$

Pin

Leave a Reply

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