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250 Answers to Exercises and Solutions to Problems

4.9. p3+8q2 = 0. If y = 2x, the latter equation becomes y4 = (~~+6y-18)~.

4.10. (a) 4 = ~12, x = 2acos r/6 = I/%.

(b) 4 = 0, x = 2.

4.11. (b) us + 2pu4 + (p2 - 4r)u2 - q2 = 0.

(d) The equation for u2 is us + 2u4 + 13u2 - 16 = 0, which leads to u = 1

and the factorization

t4 + t2 + 4t - 3 = (P + t - l)(? -t + 3).

The equation for u2 is u6 - 4u4 + 16u2 - 64 = 0 which leads to u = 2 and

the factorization

t4 - 2t2 + 8t - 3 = (P + 2t - l)(P - 2t + 3).

4.12. (b) Setting the discriminant of the quadratic equal to zero yields
2u3 - qu2 + 2(pr - s)u + (qs - p2s - r2) = 0.
(c) The first equation is equivalent to (t2 + 1)2 = (t - 2)2.
For the second equation, u is given by 2u3 + 5u2 - 4u - 7 = 0, and we can
rewrite the equation as (t2 - t - 1)2 = 4(t - 1)2.

4.13. (c) A reciprocal polynomial of odd degree can be written in the form
u(x+~ + 1) + bx(x2k-1 + 1) + CX~(X-~ + 1) +... = 0. Noting that when
m is odd we have

xm + 1 = (x + l)(xm-l - 2m-2 + Zm-3 -... - 2 + l),

we obtain the result.

4.14. (b) Prove by induction on m. If the result is true for m 5 r, then

xr+l + x-(‘+l) = qxr + x-‘) - (xr-l + x-(4),

whence the left side is a polynomial of degree r + 1.
(c) Since x = 0 is not a solution of the equation, we can replace it by the
equivalent
u(xk + x-~) + b(x’-’ + x-tk-‘)) + ... = 0.

A change to the variable t gives an equation of degree k in t. Having found
values oft, we can then solve equations of the form x2 - tx + 1 = 0 for x.

4.15. (a) t = l/2, -3.
(b) 2x4 + 5x3 + x2 + 5x + 2 = (2x2 - x + 2)(z2 + 3x + 1).

4.17. (c) x = u + v leads to a quasi-reciprocal equation in u provided that
4v3 + v2 - 6v + 1 = 0. Taking v = 1 leads to the equation u4 + 4u3 + 9u2 +
8u+4=0.Lett=u+2/u.Thent2+4t+5=0,whencet=-2fi.The
equation u2 + (2 f i)u + 2 = 0 can be solved for u (see Exercise 3.10) and
then x = u + 1.