Menu Close

Solve-for-x-C-x-3-4-3i-x-2-51-49i-x-442-170i-0-




Question Number 204999 by Frix last updated on 05/Mar/24
Solve for x∈C  x^3 +(4−3i)x^2 −(51+49i)x−442+170i=0
$$\mathrm{Solve}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$$${x}^{\mathrm{3}} +\left(\mathrm{4}−\mathrm{3i}\right){x}^{\mathrm{2}} −\left(\mathrm{51}+\mathrm{49i}\right){x}−\mathrm{442}+\mathrm{170i}=\mathrm{0} \\ $$
Answered by Skabetix last updated on 05/Mar/24
S={ −9−4i ; −3+5i ; 8+2i }?
$${S}=\left\{\:−\mathrm{9}−\mathrm{4}{i}\:;\:−\mathrm{3}+\mathrm{5}{i}\:;\:\mathrm{8}+\mathrm{2}{i}\:\right\}? \\ $$$$ \\ $$
Commented by Frix last updated on 05/Mar/24
Your method?
$$\mathrm{Your}\:\mathrm{method}? \\ $$
Commented by TonyCWX08 last updated on 06/Mar/24
Real Answer:  −8.651+6.543i  −4.331−2.06i  8.982−1.483i    Method: Wolfram Alpha
$${Real}\:{Answer}: \\ $$$$−\mathrm{8}.\mathrm{651}+\mathrm{6}.\mathrm{543}{i} \\ $$$$−\mathrm{4}.\mathrm{331}−\mathrm{2}.\mathrm{06}{i} \\ $$$$\mathrm{8}.\mathrm{982}−\mathrm{1}.\mathrm{483}{i} \\ $$$$ \\ $$$${Method}:\:{Wolfram}\:{Alpha} \\ $$
Commented by Frix last updated on 06/Mar/24
Wolfram Alpha is wrong. It′s not  possible to use Cardano′s Solution in this  case.
$$\mathrm{Wolfram}\:\mathrm{Alpha}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{It}'\mathrm{s}\:\mathrm{not} \\ $$$$\mathrm{possible}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{Solution}\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{case}. \\ $$
Commented by TonyCWX08 last updated on 08/Mar/24
Are you sure??  How you know Wolfram Alpha ude Cardano′s Formula??
$${Are}\:{you}\:{sure}?? \\ $$$${How}\:{you}\:{know}\:{Wolfram}\:{Alpha}\:{ude}\:{Cardano}'{s}\:{Formula}?? \\ $$
Commented by Frix last updated on 08/Mar/24
Because your answer is wrong. Test your  solutions by inserting.
$$\mathrm{Because}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{Test}\:\mathrm{your} \\ $$$$\mathrm{solutions}\:\mathrm{by}\:\mathrm{inserting}. \\ $$
Answered by Rasheed.Sindhi last updated on 08/Mar/24
let x=a+bi  ▶(a+bi)^3 +(4−3i)(a+bi)^2 −(51+49i)(a+bi)−442+170i=0  ▶a^3 +3a^2 bi−3ab^2 −b^3 i     +(4−3i)(a^2 +2abi−b^2 )            −(51a−49b+49ai+51bi)                                          −442+170i=0  ▶a^3 +3a^2 bi−3ab^2 −b^3 i    +4a^2 +8abi−4b^2 −3a^2 i+6ab+3b^2 i    −51a+49b−49ai−51bi    −442+170i=0  ▶a^3 −3ab^2 +4a^2 −4b^2 +6ab−51a+49b−442=0_(&^  )      3a^2 b−b^3 +8ab−3a^2 +3b^2 −49a−51b+170=0  ▶−15ab^2 −20b^2 +30ab+245b+5a^3 +20a^2 −255a−2210=0      −13b^3 +39b^2 +39a^2 b+104ab−663b−39a^2 −637a+2210=0
$${let}\:{x}={a}+{bi} \\ $$$$\blacktriangleright\left({a}+{bi}\right)^{\mathrm{3}} +\left(\mathrm{4}−\mathrm{3}{i}\right)\left({a}+{bi}\right)^{\mathrm{2}} −\left(\mathrm{51}+\mathrm{49}{i}\right)\left({a}+{bi}\right)−\mathrm{442}+\mathrm{170}{i}=\mathrm{0} \\ $$$$\blacktriangleright{a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {bi}−\mathrm{3}{ab}^{\mathrm{2}} −{b}^{\mathrm{3}} {i} \\ $$$$\:\:\:+\left(\mathrm{4}−\mathrm{3}{i}\right)\left({a}^{\mathrm{2}} +\mathrm{2}{abi}−{b}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:−\left(\mathrm{51}{a}−\mathrm{49}{b}+\mathrm{49}{ai}+\mathrm{51}{bi}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{442}+\mathrm{170}{i}=\mathrm{0} \\ $$$$\blacktriangleright{a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {bi}−\mathrm{3}{ab}^{\mathrm{2}} −{b}^{\mathrm{3}} {i} \\ $$$$\:\:+\mathrm{4}{a}^{\mathrm{2}} +\mathrm{8}{abi}−\mathrm{4}{b}^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} {i}+\mathrm{6}{ab}+\mathrm{3}{b}^{\mathrm{2}} {i} \\ $$$$\:\:−\mathrm{51}{a}+\mathrm{49}{b}−\mathrm{49}{ai}−\mathrm{51}{bi} \\ $$$$\:\:−\mathrm{442}+\mathrm{170}{i}=\mathrm{0} \\ $$$$\blacktriangleright\underset{\overset{\:} {\&}} {{a}^{\mathrm{3}} −\mathrm{3}{ab}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} +\mathrm{6}{ab}−\mathrm{51}{a}+\mathrm{49}{b}−\mathrm{442}=\mathrm{0}} \\ $$$$\:\:\:\mathrm{3}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} +\mathrm{8}{ab}−\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} −\mathrm{49}{a}−\mathrm{51}{b}+\mathrm{170}=\mathrm{0} \\ $$$$\blacktriangleright−\mathrm{15}{ab}^{\mathrm{2}} −\mathrm{20}{b}^{\mathrm{2}} +\mathrm{30}{ab}+\mathrm{245}{b}+\mathrm{5}{a}^{\mathrm{3}} +\mathrm{20}{a}^{\mathrm{2}} −\mathrm{255}{a}−\mathrm{2210}=\mathrm{0} \\ $$$$\:\:\:\:−\mathrm{13}{b}^{\mathrm{3}} +\mathrm{39}{b}^{\mathrm{2}} +\mathrm{39}{a}^{\mathrm{2}} {b}+\mathrm{104}{ab}−\mathrm{663}{b}−\mathrm{39}{a}^{\mathrm{2}} −\mathrm{637}{a}+\mathrm{2210}=\mathrm{0} \\ $$

Leave a Reply

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