Lowering the exponents

Algebra Level 3

{ x 2 + y 2 = 2 x 3 + y 3 = − 2 \large{\begin{cases}x^2+y^2=2 \\ x^3+y^3 = -2 \end{cases}}

If x x and y y are number satisfying the system of equations above, find the value(s) of x + y x+y .

(A) : − 2 -2
(B) : 1 + 3 1 +\sqrt3
(C) : 1 − 3 1 -\sqrt3

Only (A) All of (A), (B) and (C) Only (C) Only (B)

Edgar de Asis Jr. by Edgar de Asis Jr. , 25, Philippines

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1 solution

From the first given equation

( x + y ) 2 = 2 + 2 x y x y = ( x + y ) 2 − 2 2 { (x+y) }^{ 2 }=2+2xy\\ xy=\frac { { (x+y) }^{ 2 }-2 }{ 2 } .

Factoring the second equation and replacing the first one

( x + y ) ( 2 − x y ) = − 2 ( x + y ) ( 2 − ( x + y ) 2 − 2 2 ) = − 2 ( x + y ) ( 6 − ( x + y ) 2 ) = − 4 6 ( x + y ) − ( x + y ) 3 + 4 = 0 − ( x + y + 2 ) ( ( x + y ) 2 − 2 ( x + y ) − 2 ) = 0 (x+y)(2-xy)=-2\\ (x+y)\left( 2-\frac { { (x+y) }^{ 2 }-2 }{ 2 } \right) =-2\\ (x+y)(6-{ (x+y) }^{ 2 })=-4\\ 6(x+y)-{ (x+y })^{ 3 }+4=0\\ -(x+y+2)({ (x+y) }^{ 2 }-2(x+y)-2)=0

Therefore, x + y = − 2 , 1 − 3 , 1 + 3 x+y=-2,\quad 1-\sqrt { 3 } ,\quad 1+\sqrt { 3 } .

I think there's a typo in the options.

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But the options given are:

{ − 2 − 1 − 3 3 − 1 \begin {cases} - 2 \\ - 1-\sqrt 3 \\ \sqrt 3-1\end {cases}

Therefore I answered - 2.

Chew-Seong Cheong - 5 years ago

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I think the author wrote incorrectly the other options... clearly, he wanted the answer to be "all the options".

Mateo Matijasevick - 5 years ago

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