Problem:
Real numbers x and y satisfy x+y=4 and x⋅y=−2. What is the value of
x+y2x3+x2y3+y?
Answer Choices:
A. 360
B. 400
C. 420
D. 440
E. 480
Solution:
x+y2x3+x2y3+y=x+y2x3+y+x2y3=x2x3+x2y3+y2y3+y2x3
Continuing to combine
x2x3+y3+y2x3+y3=x2y2(x2+y2)(x3+y3)=x2y2(x2+y2)(x+y)(x2−xy+y2)
From the givens, it can be concluded that x2y2=4. Also,
(x+y)2=x2+2xy+y2=16
This means that
x2+y2=20
Substituting this information into
x2y2(x2+y2)(x+y)(x2−xy+y2)
we have
4(20)(4)(22)=20⋅22= (D) 440.
OR
As above, we need to calculate
x2y2(x2+y2)(x3+y3)
Note that x,y, are the roots of x2−4x−2 and so x3=4x2+2x and y3=4y2+2y. Thus
x3+y3=4(x2+y2)+2(x+y)=4(20)+2(4)=88
where x2+y2=20 and x2y2=4 as in the previous solution. Thus the answer is
4(20)(88)=(D) 440.
Note:(x2+y2=(x+y)2−2xy=20,andx2y2=(xy)2=4)
The problems on this page are the property of the MAA's American Mathematics Competitions