Find one ordered pair $(x,y)$ of real numbers such that $x + y = 10$ and $x^3 + y^3 = 162 + x^2 + y^2.$

sandwich Nov 10, 2023

#1**0 **

x + y = 10

x^3 + y^3 = 162 + x^2 + y^2

x^3 + y^3 factors into (x+y)(x^2+xy+y^2)

we know x+y is 10 from the first equation so we plug that in

so now we have:

10(x^2+xy+y^2) = 162+x^2+y^2

expanding gives us

10x^2+10xy+10y^2 = 162+x^2+y^2

combine like terms

9x^2 + 9y^2 + 10xy - 162 = 0

divdie both sides by 9

x^2 + y^2 + 9/10xy - 18 = 0

Since x+y = 10, (x+y)^2 = 100 and x^2+y^2 = 100 - 2xy

plugging that in gives

100-2xy+9/10xy-18=0

combine like terms

82-11/10xy=0

11/10xy = 82

xy = 820/11

now knowing the sum and product of x and y, we can create quadratic satisfying these properties using vieta's and solve

x+y = -b/a

xy = c/a

let a = 1 for convenience

we get

n^2 -10n + 820/11 = 0

solving gives

x and y = **5 + i(sqrt(545/11)) and 5 - i(sqrt(545/11))**

LurpDaDerp Nov 10, 2023

#2**0 **

If we factor the left side of the second equation, we get the equation:

\((x+y)(x^2+xy+y^2)=162+x^2+y^2\)

Since the first equation gives x+y=10, the second equation now becomes:

\(10(x^2+xy+y^2)=162+x^2+y^2 \)

\(10x^2+10xy+10y^2= 162+x^2+y^2\)

\(9x^2+9y^2+10xy=162\)

Now substitute the first equation into the second

\(9(10-y)^2+9y^2+10(10-y)y=162\)

\(9(100-20y+y^2)+9y^2+100y-10y^2=162\)

\(900-180y+9y^2+9y^2+100y-10y^2=162\)

\(8y^2-80y+900=162\)

\(8y^2-80y+738=0\)

However, putting this into the quadratic formula, we find that the discriminant is negative, so therefore there are no real solutions.

Where did you find this problem?

Erwus Nov 10, 2023