5.2 ALGEBRAIC MANIPULATION REVISITED 151equation x^4 + x^3 + x^2 + x + 1 = 0 to two quadratic equations. Here are a few more
examples.
Example 5.2. 16 (AIME 198 3) What is the product of the real roots of the equation
y. + 1 8 x+30 = 2Jx^2 + 1 8 x+45?
Solution: This is not a very hard problem. The only real obstacle to immediate
solution is the fact that there is a square root. The first thing to try, then, is to eliminate
this obstacle by boldly substituting
y = Jx^2 +^18 x+4 (^5).
Notice that if x is real, then y must be non-negative. The equation immediately simpli
fies to
i- 1 5 =2y,
which factors nicely into (y -5)(y+ 3) = O. Reject the root y = - 3 (since y must be
non-negative); substituting the root y = 5 back into the original substitution yields
x^2 + 1 8 x + 45 = 5^2
or
y. + 1 8 x+2 0 = O.
Hence the product of the roots is 20, using the relationship between zeros and coeffi
cients formula (see page 168 ). •
Example 5.2. 17 (AIME 198 6) Simplify
( Vs + J6 +^0 ) ( Vs + J6 -^0 ) ( Vs -J6 +^0 ) ( - Vs + J6 +^0 ).
We could mUltiply out all the terms, but it would take a long time, and we'd probably
make a mistake. We need a strategy. If this expression is to simplify, we will probably
be able to eliminate radicals. If we multiply any two terms, we can use the difference
of two squares formula (5.2.6) and get expressions which contain only one radical. For
example, the product of the first and second terms is
( Vs + J6 +^0 ) ( Vs + J6 - 0) = ( Vs + J6)
2
_ ( 0 )
2
Likewise, the product of the last two terms is
=5+6+2v'30-7
=4+2v'30.
( 0 + (Vs-J6)) ( 0 - (Vs -J6)) = 7 -(5-2v'30+6) = -4+2v'30.
The final product, then, is
(4+2v'30) ( -4+2v'30) = 4· 30 -16 = 10 4. •