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


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

$Suppose that the sum of the square of$ $complex numbers x or y is 7, and the$ $sum of their cubes is 10 . Find the largest$ $true value of the sum x + y that satisfies$ $these conditions .$ $A) 4 B) 5 C) 6 D) 7 E) 8$
Commented by mr W last updated on 08/Jan/23

$w h a t i s m e a n t w i t h ‘ ‘ t r u e {v a l u e}^{″} ?$
Commented by Shrinava last updated on 08/Jan/23

$dear professor, real price$
Commented by mr W last updated on 08/Jan/23

$w h a t i s t h e n m e a n t w i t h ‘ ‘ r e a l {p r i c e}^{″} ?$ $i f y o u t r a n s l a t e f r o m y o u r n a t i v e$ $l a n g u a g e, p l e a s e g i v e a t b e s t a n$ $e x a m p l e s h o w i n g w h a t i s c o n c r e t e l y$ $m e a n t . e . g . w h a t i s t h e ‘ ‘ r e a l {p r i c e}^{″}$ $f r o m t h e n u m b e r 2 + 3 i ?$
Commented by mr W last updated on 08/Jan/23

$i g u e s s y o u m e a n t h e a b s o l u t e v a l u e$ $o f a c o m p l e x n u m b e r .$
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

$yes dear professor mrW$
	Answered by Frix last updated on 08/Jan/23
	$We must solve in C (1) x^2 +y^2 =7 ⇒ y=±(√(7−x^2 )) (2) x^3 +y^3 =10 ⇔ x^3 ±(7−x^2 )^(3/2) =10 ⇒ ±(7−x^2 )^(3/2) =10−x^3 Squaring [might introduce false solutions] and transforming ⇒ x^6 −((21x^4 )/2)−10x^3 +((147x^2 )/2)−((243)/2)=0 I used software to find x=((1±(√(13)))/2)∨x=−(5/2)±((√(11))/2)i∨x=2±((√2)/2)i Testing leads to these pairs: x=((1±(√(13)))/2)∧y=((1∓(√(13)))/2) x=−(5/2)±((√(11))/2)i∧y=−(5/2)∓((√(11))/2)i x=2±((√2)/2)i∧y=2∓((√2)/2)i ⇒ x+y∈{−5, 1, 4}$
	$We must solve in C$ $(1) x^{2} + y^{2} = 7 \Rightarrow y = \pm \sqrt{7 - x^{2}}$ $(2) x^{3} + y^{3} = 10 \Leftrightarrow x^{3} \pm {(7 - x^{2})}^{\frac{3}{2}} = 10 \Rightarrow$ $\pm {(7 - x^{2})}^{\frac{3}{2}} = 10 - x^{3}$ $Squaring [might introduce false solutions]$ $and transforming \Rightarrow$ $x^{6} - \frac{21 x^{4}}{2} - 10 x^{3} + \frac{147 x^{2}}{2} - \frac{243}{2} = 0$ $I used software to find$ $x = \frac{1 \pm \sqrt{13}}{2} \lor x = - \frac{5}{2} \pm \frac{\sqrt{11}}{2} i \lor x = 2 \pm \frac{\sqrt{2}}{2} i$ $Testing leads to these pairs :$ $x = \frac{1 \pm \sqrt{13}}{2} \land y = \frac{1 \mp \sqrt{13}}{2}$ $x = - \frac{5}{2} \pm \frac{\sqrt{11}}{2} i \land y = - \frac{5}{2} \mp \frac{\sqrt{11}}{2} i$ $x = 2 \pm \frac{\sqrt{2}}{2} i \land y = 2 \mp \frac{\sqrt{2}}{2} i$ $\Rightarrow$ $x + y \in {- 5, 1, 4}$
	Answered by Frix last updated on 08/Jan/23
	$Much easier to solve: x=u−v y=u+v ⇒ (1) 2u^2 +2v^2 =7 ⇒ v^2 =(7/2)−u^2 (2) 2u^3 +6uv^2 =10 ⇒ v^2 =((5−u^3 )/(3u)) ⇒ (7/2)−u^2 =((5−u^3 )/(3u)) ⇔ u^3 −((21u)/4)+(5/2)=0 ⇒ u=−(5/2)∨u=(1/2)∨u=2 We don′t even need the values of v because x+y=2u∈{−5, 1, 4}$
	$Much easier to solve :$ $x = u - v$ $y = u + v$ $\Rightarrow$ $(1) 2 u^{2} + 2 v^{2} = 7 \Rightarrow v^{2} = \frac{7}{2} - u^{2}$ $(2) 2 u^{3} + 6 {u v}^{2} = 10 \Rightarrow v^{2} = \frac{5 - u^{3}}{3 u}$ $\Rightarrow$ $\frac{7}{2} - u^{2} = \frac{5 - u^{3}}{3 u}$ $\Leftrightarrow$ $u^{3} - \frac{21 u}{4} + \frac{5}{2} = 0$ $\Rightarrow u = - \frac{5}{2} \lor u = \frac{1}{2} \lor u = 2$ $We {don}^{'} t even need the values of v because$ $x + y = 2 u \in {- 5, 1, 4}$
	Answered by mr W last updated on 08/Jan/23

	$x^{2} + y^{2} = 7$ ${(x + y)}^{2} - 2 x y = 7 \Rightarrow x y = \frac{{(x + y)}^{2} - 7}{2}$ $x^{3} + y^{3} = (x + y) (x^{2} + y^{2} - x y) = 10$ $(x + y) (7 - \frac{{(x + y)}^{2} - 7}{2}) = 10$ ${(x + y)}^{3} - 21 (x + y) + 20 = 0$ $(x + y - 1) (x + y - 4) (x + y + 5) = 0$ $\Rightarrow x + y = 1 o r 4 o r - 5$ $∣ x + y ∣= 1 o r 4 o r 5$ $∣ x + y ∣_{m a x} = 5$ $\Rightarrow a n s w e r (B)$
	Commented by Shrinava last updated on 09/Jan/23

	$perfect dear professor, thank you$
	Commented by mr W last updated on 09/Jan/23

	$y o u s h o u l d a l s o t h a n k o t h e r p e o p l e$ $w h o s o l v e d y o u r q u e s t i o n e v e n b e f o r e$ $m e!$
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