#### Question

If x+y = 1, x^4+y^4 = 881, find the value of xy

Question

If x+y = 1, x^4+y^4 = 881, find the value of xy

\begin{aligned} x^4+y^4 &= (x^2+y^2)^2-2x^2y^2 \\ &=[(x+y)^2-2xy]^2-2x^2y^2 \\ & = 881\end{aligned}

Substitution of the given condition x+y = 1 results in a quadratic equation in terms of xy

(1-2xy)^2-2(xy)^2 = 881

(xy)^2-2xy-440 = 0

Factoring the LHS

(xy-22)(xy+20) = 0

Therefore

xy = 22 \text{ or } xy = -20