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Question Number 184555 by Shrinava last updated on 08/Jan/23

Suppose that the sum of the square of complex numbers x or y is 7 , and the sum of their cubes is 10. Find the largest true value of the sum x+y that satisfies these conditions. A)4 B)5 C)6 D)7 E)8

$$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of} \\ $$$$\mathrm{complex}\:\mathrm{numbers}\:\:\boldsymbol{\mathrm{x}}\:\mathrm{or}\:\boldsymbol{\mathrm{y}}\:\mathrm{is}\:\mathrm{7}\:,\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{cubes}\:\mathrm{is}\:\mathrm{10}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{largest} \\ $$$$\mathrm{true}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sum}\:\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\:\:\mathrm{that}\:\mathrm{satisfies} \\ $$$$\mathrm{these}\:\mathrm{conditions}. \\ $$$$\left.\mathrm{A}\left.\right)\left.\mathrm{4}\left.\:\left.\:\:\mathrm{B}\right)\mathrm{5}\:\:\:\mathrm{C}\right)\mathrm{6}\:\:\:\mathrm{D}\right)\mathrm{7}\:\:\:\mathrm{E}\right)\mathrm{8} \\ $$

Commented by mr W last updated on 08/Jan/23

what is meant with “true value”?

$${what}\:{is}\:{meant}\:{with}\:``{true}\:{value}''? \\ $$

Commented by Shrinava last updated on 08/Jan/23

dear professor, real price

$$\mathrm{dear}\:\mathrm{professor},\:\mathrm{real}\:\mathrm{price} \\ $$

Commented by mr W last updated on 08/Jan/23

what is then meant with “real price”? if you translate from your native language, please give at best an example showing what is concretely meant. e.g. what is the “real price” from the number 2+3i ?

$${what}\:{is}\:{then}\:{meant}\:{with}\:``{real}\:{price}''? \\ $$$${if}\:{you}\:{translate}\:{from}\:{your}\:{native} \\ $$$${language},\:{please}\:{give}\:{at}\:{best}\:{an}\: \\ $$$${example}\:{showing}\:{what}\:{is}\:{concretely} \\ $$$${meant}.\:{e}.{g}.\:{what}\:{is}\:{the}\:``{real}\:{price}''\: \\ $$$${from}\:{the}\:{number}\:\mathrm{2}+\mathrm{3}{i}\:? \\ $$

Commented by mr W last updated on 08/Jan/23

i guess you mean the absolute value of a complex number.

$${i}\:{guess}\:{you}\:{mean}\:{the}\:{absolute}\:{value} \\ $$$${of}\:{a}\:{complex}\:{number}. \\ $$

Commented by JDamian last updated on 08/Jan/23

I think this kind of misunderstanding can be avoided by using more expressions in math language instead of long translated clauses from native languages

Commented by Shrinava last updated on 08/Jan/23

yes dear professor mrW

$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{mrW} \\ $$

Answered by Frix last updated on 08/Jan/23

We must solve in C (1) x^2 +y^2 =7 ⇒ y=±(√(7−x^2 )) (2) x^3 +y^3 =10 ⇔ x^3 ±(7−x^2 )^(3/2) =10 ⇒ ±(7−x^2 )^(3/2) =10−x^3 Squaring [might introduce false solutions] and transforming ⇒ x^6 −((21x^4 )/2)−10x^3 +((147x^2 )/2)−((243)/2)=0 I used software to find x=((1±(√(13)))/2)∨x=−(5/2)±((√(11))/2)i∨x=2±((√2)/2)i Testing leads to these pairs: x=((1±(√(13)))/2)∧y=((1∓(√(13)))/2) x=−(5/2)±((√(11))/2)i∧y=−(5/2)∓((√(11))/2)i x=2±((√2)/2)i∧y=2∓((√2)/2)i ⇒ x+y∈{−5, 1, 4}

$$\mathrm{We}\:\mathrm{must}\:\mathrm{solve}\:\mathrm{in}\:\mathbb{C} \\ $$$$\left(\mathrm{1}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{7}\:\Rightarrow\:{y}=\pm\sqrt{\mathrm{7}−{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{10}\:\Leftrightarrow\:{x}^{\mathrm{3}} \pm\left(\mathrm{7}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{10}\:\Rightarrow \\ $$$$\pm\left(\mathrm{7}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{10}−{x}^{\mathrm{3}} \\ $$$$\mathrm{Squaring}\:\left[\mathrm{might}\:\mathrm{introduce}\:\mathrm{false}\:\mathrm{solutions}\right] \\ $$$$\mathrm{and}\:\mathrm{transforming}\:\Rightarrow \\ $$$${x}^{\mathrm{6}} −\frac{\mathrm{21}{x}^{\mathrm{4}} }{\mathrm{2}}−\mathrm{10}{x}^{\mathrm{3}} +\frac{\mathrm{147}{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{243}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{used}\:\mathrm{software}\:\mathrm{to}\:\mathrm{find} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{13}}}{\mathrm{2}}\vee{x}=−\frac{\mathrm{5}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\vee{x}=\mathrm{2}\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{Testing}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{these}\:\mathrm{pairs}: \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{13}}}{\mathrm{2}}\wedge{y}=\frac{\mathrm{1}\mp\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${x}=−\frac{\mathrm{5}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\wedge{y}=−\frac{\mathrm{5}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i} \\ $$$${x}=\mathrm{2}\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\wedge{y}=\mathrm{2}\mp\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${x}+{y}\in\left\{−\mathrm{5},\:\mathrm{1},\:\mathrm{4}\right\} \\ $$

Answered by Frix last updated on 08/Jan/23

Much easier to solve: x=u−v y=u+v ⇒ (1) 2u^2 +2v^2 =7 ⇒ v^2 =(7/2)−u^2 (2) 2u^3 +6uv^2 =10 ⇒ v^2 =((5−u^3 )/(3u)) ⇒ (7/2)−u^2 =((5−u^3 )/(3u)) ⇔ u^3 −((21u)/4)+(5/2)=0 ⇒ u=−(5/2)∨u=(1/2)∨u=2 We don′t even need the values of v because x+y=2u∈{−5, 1, 4}

$$\mathrm{Much}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{solve}: \\ $$$${x}={u}−{v} \\ $$$${y}={u}+{v} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{v}^{\mathrm{2}} =\mathrm{7}\:\Rightarrow\:{v}^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{2}}−{u}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2}{u}^{\mathrm{3}} +\mathrm{6}{uv}^{\mathrm{2}} =\mathrm{10}\:\Rightarrow\:{v}^{\mathrm{2}} =\frac{\mathrm{5}−{u}^{\mathrm{3}} }{\mathrm{3}{u}} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}−{u}^{\mathrm{2}} =\frac{\mathrm{5}−{u}^{\mathrm{3}} }{\mathrm{3}{u}} \\ $$$$\Leftrightarrow \\ $$$${u}^{\mathrm{3}} −\frac{\mathrm{21}{u}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:{u}=−\frac{\mathrm{5}}{\mathrm{2}}\vee{u}=\frac{\mathrm{1}}{\mathrm{2}}\vee{u}=\mathrm{2} \\ $$$$\mathrm{We}\:\mathrm{don}'\mathrm{t}\:\mathrm{even}\:\mathrm{need}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{v}\:\mathrm{because} \\ $$$${x}+{y}=\mathrm{2}{u}\in\left\{−\mathrm{5},\:\mathrm{1},\:\mathrm{4}\right\} \\ $$

Answered by mr W last updated on 08/Jan/23

x^2 +y^2 =7 (x+y)^2 −2xy=7 ⇒xy=(((x+y)^2 −7)/2) x^3 +y^3 =(x+y)(x^2 +y^2 −xy)=10 (x+y)(7−(((x+y)^2 −7)/2))=10 (x+y)^3 −21(x+y)+20=0 (x+y−1)(x+y−4)(x+y+5)=0 ⇒x+y=1 or 4 or −5 ∣x+y∣=1 or 4 or 5 ∣x+y∣_(max) =5 ⇒answer (B)

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{7} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{7}\:\Rightarrow{xy}=\frac{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{7}}{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}\right)=\mathrm{10} \\ $$$$\left({x}+{y}\right)\left(\mathrm{7}−\frac{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{7}}{\mathrm{2}}\right)=\mathrm{10} \\ $$$$\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{21}\left({x}+{y}\right)+\mathrm{20}=\mathrm{0} \\ $$$$\left({x}+{y}−\mathrm{1}\right)\left({x}+{y}−\mathrm{4}\right)\left({x}+{y}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}+{y}=\mathrm{1}\:{or}\:\mathrm{4}\:{or}\:−\mathrm{5} \\ $$$$\mid{x}+{y}\mid=\mathrm{1}\:{or}\:\mathrm{4}\:{or}\:\mathrm{5} \\ $$$$\mid{x}+{y}\mid_{{max}} =\mathrm{5} \\ $$$$\Rightarrow{answer}\:\left({B}\right) \\ $$

Commented by Shrinava last updated on 09/Jan/23

perfect dear professor, thank you

$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor},\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mr W last updated on 09/Jan/23

you should also thank other people who solved your question even before me !

$${you}\:{should}\:{also}\:{thank}\:{other}\:{people}\: \\ $$$${who}\:{solved}\:{your}\:{question}\:{even}\:{before}\: \\ $$$${me}\:! \\ $$

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