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

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

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

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

$$whatismeantwith“true{value}^{″}?$$

Commented by Shrinava last updated on 08/Jan/23

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

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

$$whatisthenmeantwith“real{price}^{″}?$$$$ifyoutranslatefromyournative$$$$language,pleasegiveatbestan$$$$exampleshowingwhatisconcretely$$$$meant.e.g.whatisthe“real{price}^{″}$$$$fromthenumber2+3i?$$

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

$$iguessyoumeantheabsolutevalue$$$$ofacomplexnumber.$$

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

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

Answered by Frix last updated on 08/Jan/23

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

Answered by Frix last updated on 08/Jan/23

$$\mathrm{Much}\mathrm{easier}\mathrm{to}\mathrm{solve}:$$$$x=u-v$$$$y=u+v$$$$\Rightarrow$$$$\left(1\right)2u^2+2v^2=7\Rightarrow v^2=\frac{7}{2}-u^2$$$$\left(2\right)2u^3+6{uv}^2=10\Rightarrow v^2=\frac{5-u^3}{3u}$$$$\Rightarrow$$$$\frac{7}{2}-u^2=\frac{5-u^3}{3u}$$$$\iff$$$$u^3-\frac{21u}{4}+\frac{5}{2}=0$$$$\Rightarrow u=-\frac{5}{2}\vee u=\frac{1}{2}\vee u=2$$$$\mathrm{We}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{even}\mathrm{need}\mathrm{the}\mathrm{values}\mathrm{of}v\mathrm{because}$$$$x+y=2u\in \{-5,1,4\}$$

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

$$x^2+y^2=7$$$${(x+y)}^2-2xy=7\Rightarrow xy=\frac{{(x+y)}^2-7}{2}$$$$x^3+y^3=(x+y)(x^2+y^2-xy)=10$$$$(x+y)(7-\frac{{(x+y)}^2-7}{2})=10$$$${(x+y)}^3-21(x+y)+20=0$$$$(x+y-1)(x+y-4)(x+y+5)=0$$$$\Rightarrow x+y=1or4or-5$$$$\mid x+y\mid =1or4or5$$$$\mid x+y{\mid }_{max}=5$$$$\Rightarrow answer\left(B\right)$$

Commented by Shrinava last updated on 09/Jan/23

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

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

$$youshouldalsothankotherpeople$$$$whosolvedyourquestionevenbefore$$$$me!$$

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