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308 Answers to Exercises and Solutions to Problems

For there to be infinitely many solutions, it is necessary that a = -2c

and b = c2. Suppose these conditions hold. Then the equation reduces to

\/ (A--cpc-t/F.

If c < 0, the equation has no solution. If c 2 0, then it is satisfied by all I
for which 0 < z 5 c2. Hence there are infinitely many solutions iff c > 0
and a = -2c, b = c.
7.6. Square both sides:

2x+2dm=k2.

There are two possible cases to consider: (1) z > 1; (2) x 5 1. If x > 1,
then the equation reduces to 42: = k2 + 2. A necessary condition for a
solution x > 1 is that k > a. If 4 > 2, then it can be checked that
x = (k2 + 2)/4 indeed satisfies the equation. If x 5 1, then the equation
reduces to 2x+2( l-x) = k2, or 2 = k2. Hence, for a solution, it is necessary
that k = 4. On the other hand, let 2 be any number with l/2 5 x 5 1.
Then it is readily checked that the square of the left side is 2.
Thus, if k2 > 2, then x = (k2+2)/4 satisfies the equation, while if k2 = 2,
any x with 1 5 2x 5 2 satisfies the equation.
7.7. Squaring both sides of the equation and collecting terms gives:

2ub - y = 2&j73,

where y = x[x - (u + b)]. Sq uaring again leads to y2 = &by, whence y =^0
or y = 8ub. The solution y = 8ub is extraneous (since 2ub - y would be
negative). Hence x = 0 or 2: = a + b. The first of these is extraneous, but
x = a + b is a valid solution.
7.8. Let y = x2 + 5x. The equation becomes

(Y + 4)(~ + 6) + (Y - 6) = 0,

whose solutions are y = -2 and y = -9. We can now form two quadratic

equations for the four values of x.

7.9. (a) p(t) = t + u(t - u)(t - v) (by the Factor Theorem)

=a p(t) - u = (t - u)[u(t - v) + 11;

p(t) - v = (t - v)[a(t - U) + l]

* p(p(t)) - t = p(p(t)) - p(t) + p(t) - t

= ah(t) - u)(p(t) - v) + (t - u)(t - v)]

= u(t - u)(t - v)[u2(t - u)(t - v) + u(2t - U - v) + 1 + l]

= u(t - u)(t - v)[u(p(t) - t) + u(2t) + (b - 1) + 21

= a(t - u)(t - v)[u2t2 + u(b + 1)t + ac + b + 11.