Question Number 104973 by malwaan last updated on 25/Jul/20 | ||
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$$\mathrm{2}\sqrt{\mathrm{3}}\:+\:\boldsymbol{{i}}\:\:\:{is}\:{a}\:{cubic}\:{root}\:{for} \\ $$$$\mathrm{18}\sqrt{\mathrm{3}}\:+\:\mathrm{35}\boldsymbol{{i}}\: \\ $$$$\boldsymbol{{find}}\:\:\boldsymbol{{the}}\:\boldsymbol{{other}}\:\mathrm{2}\:\boldsymbol{{cubic}}\:\boldsymbol{{roots}} \\ $$ | ||
Commented by malwaan last updated on 25/Jul/20 | ||
شكرا جزيلا سيدي الفاضل واذا كان لديك طريقة او لوحة مفاتيح لكتابة المعادلات والنهايات والتكاملات وغيرها باللغة العربية فاتمنى تساعدني مع فائق احترامي محمد علوان اليمن | ||
Answered by Rasheed.Sindhi last updated on 25/Jul/20 | ||
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$${Let}\:{a}\:{is}\:{a}\:{given}\:{cuberoot}\:{of}\:{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({say}\right) \\ $$$$\therefore\:{x}^{\mathrm{3}} ={a}^{\mathrm{3}} \Rightarrow{x}^{\mathrm{3}} −{a}^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow\left({x}−{a}\right)\left({x}^{\mathrm{2}} +{ax}+{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{x}−{a}=\mathrm{0}\:\vee\:{x}^{\mathrm{2}} +{ax}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:{x}={a}\left({given}\right)\:\vee\:{x}=\frac{−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{−{a}\pm{ai}\sqrt{\mathrm{3}}}{\mathrm{2}}={a}\left(\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}={a}\omega,{a}\omega^{\mathrm{2}} \\ $$$${a}=\mathrm{2}\sqrt{\mathrm{3}}\:+\:\boldsymbol{{i}}\rightarrow \\ $$$${x}=\left(\mathrm{2}\sqrt{\mathrm{3}}\:+\:\boldsymbol{{i}}\right)\omega\:,\:\left(\mathrm{2}\sqrt{\mathrm{3}}\:+\:\boldsymbol{{i}}\right)\omega^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\left({Other}\:{two}\:{cuberoots}\right) \\ $$ | ||
Commented by malwaan last updated on 25/Jul/20 | ||
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$${wow} \\ $$$${how}\:{do}\:{you}\:{write}\:{in}\:{arabic}\:{sir}\:? \\ $$$${from}\:{this}\:{app}\:{or}\:{other}\:? \\ $$ | ||
Commented by Rasheed.Sindhi last updated on 25/Jul/20 | ||
Use the option 'plain text comment' to write with your arabic keyboard: ل^۲+۸ل+۲۲=۷ √(8)=2√(2) | ||
Commented by Rasheed.Sindhi last updated on 25/Jul/20 | ||
لا ريب يا سيدي! | ||
Commented by malwaan last updated on 25/Jul/20 | ||
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$${thank}\:{sir}\:{Rasheed} \\ $$$${and}\:{if}\:{we}\:{want}\:{the}\:{answer} \\ $$$${without}\:{using}\:\omega\:{and}\:\omega^{\mathrm{2}} \\ $$$${x}=\left(\mathrm{2}\sqrt{\mathrm{3}}\:+{i}\right)\left(\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\frac{−\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{5}{i}}{\mathrm{2}} \\ $$$${x}=\left(\mathrm{2}\sqrt{\mathrm{3}}\:+{i}\right)\left(\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\frac{−\sqrt{\mathrm{3}}−\mathrm{7}{i}}{\mathrm{2}} \\ $$ | ||