Find the complex solutions

Algebra Level 5

{ x 5 + y 5 = 33 x + y = 3 \large \begin{cases} {x^5 + y^5=33} \\ {x+y=3 } \end{cases}

If the system of equations above has two real and two complex solutions, what is the value of x 2 + y 2 x^2+y^2 when x , y x,y are both not real numbers?

Note that a + b − 1 a + b\sqrt{-1} is a complex number where a , b a,b are real numbers.


The answer is -5.

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Navin Manaswi by Navin Manaswi , 41, India

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1 solution

( x + y ) 5 = x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 x y 4 + y 5 3 5 = 33 + 5 x y ( x 3 + y 3 ) + 10 x 2 y 2 ( x + y ) 243 = 33 + 5 x y ( x + y ) ( x 2 + y 2 − x y ) + 30 x 2 y 2 210 = 15 x y ( ( x + y ) 2 − 3 x y ) + 30 x 2 y 2 70 = 5 x y ( 9 − 3 x y ) + 10 x 2 y 2 70 = 45 x y − 5 x 2 y 2 \begin{aligned} (x+y)^5 & = x^5+5x^4y+10x^3y^2+10x^2y^3+ 5 x y^4 +y^5 \\ 3^5 & = 33 +5xy\left(x^3+y^3\right) + 10x^2 y^2 \left( x+ y \right) \\ 243 & = 33 + 5xy (x+y) \left(x^2+y^2-xy \right) + 30x^2 y^2 \\ 210 & = 15xy\left((x+y)^2-3xy \right) + 30x^2 y^2 \\ 70 & = 5xy \left( 9 -3xy \right) + 10 x^2 y^2 \\ 70 & = 45xy - 5 x^2 y^2 \end{aligned}

⇒ x 2 y 2 − 9 x y + 14 = 0 ( x y − 2 ) ( x y − 7 ) = 0 ⇒ x y = 2 or 7 \begin{aligned} \Rightarrow x^2y^2 - 9xy + 14 & = 0 \\ (xy-2)(xy-7) & = 0 \\ \Rightarrow xy & = 2 \text{ or } 7 \end{aligned}

Since x + y = 3 ⇒ y = 3 − x ⇒ x y = x ( 3 − x ) = 3 x − x 2 x+y=3 \quad \Rightarrow y = 3-x\quad \Rightarrow xy = x(3-x) = 3x-x^2

3 x − x 2 = { 2 ⇒ x 2 − 3 x + 2 = ( x − 1 ) ( x − 2 ) = 0 x, y are real 7 ⇒ x 2 − 3 x + 7 = 0 x, y are not real 3x-x^2 = \begin{cases} 2 &\Rightarrow x^2 - 3x + 2 = (x-1)(x-2) = 0 & \text{x, y are real} \\ 7 &\Rightarrow x^2 - 3x + 7 = 0 & \text{x, y are not real} \end{cases}

When x y = 7 ⇒ x 2 + y 2 = ( x + y ) 2 − 2 x y = 9 − 14 = − 5 xy = 7\quad \Rightarrow x^2+y^2 = (x+y)^2 - 2xy = 9 - 14 = \boxed{-5}

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Moderator note:

This approach can be motivated by the idea of Newtons Sum, in which we only need the value of x y xy , x + y x+y and x n + y n x^n + y^n can be determined from the other 2.

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