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Solutions to Problems; Chapter (^8 377)

24. First solution. Let

f(Y) = (~-fi-mY-\/z+mY+~-~)(y+Jz+~)

= Y - lOy2 + 1.

Then

f(x-45) = x4 - 4&x3 + 20x2 - 24

= -[(3x4 - 20x2 + 24) - (x - &)4x3].

Since f(&+fi+fi) = 0, we can takep(x) = 3x4-20x2+24, q(x) = 4x3.
Second solution. Let u = a+&+&, v = &+a. Then u2-2uv+v2 =
5 and v2 = 5+2&*u2- 2uv = -2& 3 u4 - 4u3v + 4u2v2 = 24 s-
u4 - 4u3v + 4u2( 5 - u2 + 2uv) = 24 3 4u3v = 3u4 - 20u2 + 24, which yields
the result. [Solution by Jeff Higham.]
Remarks. Other possibilities are p(x) = -5x7+194x5-1520x3+3120x,
q(x) = 576 (Colin Springer) and p(x) = x5 - 24x, q(x) = x4 - 20x2 - 24
(Graham Denham). Prom this result, it can be deduced that fi+ &+ 6
must be nonrational. The problem is due to Gregg Petruno, who obtained
a more general result; consult Gregg N. Petruno, Sums of irrational square
roots are irrational. Math. Magazine 61 (1988)’ 44-45.

25. Square both sides and rearrange terms to obtain

~~J=x2+3x+2... (1)

Square again, rearrange and factor to obtain

(x2 + x-p)(x2+8x+2p+9)=0... (2)

There are four roots to equation (2):

u = (-1+ l/m)/2 b = (-l- ,/m)/2

c=--4+dm d=--4-JT?Zj

For a viable solution x to the given equation, it is necessary that all
radicals and quantities under radicals be positive:

(a) x2 f 3x + 2 = (x + 2)(x + 1) 2 0, i.e. 2 5 -2 or x > -1;

(b) x2<2p+1;

(c) 3x + p + 4 > 0;

(d) x2 +9x + 9 + 3p 10.