Question

If f(x) = r2+x - 2, whose zeroes are a & B, then find th
polynomial whose zeroes are (2a -- 1) & (28 - 1)​

Answer 1

Answer:

1.1(a). The point E can be obtained by folding the square along the bisector of

angle BAC, so that AB is folded onto the diagonal with B landing on E. If F is

the point where the bisector intersects B C, then B F folds to E F. Thus E F -1 AC

and BF = EF. Since L.EFC = 45° = L.ECF, FE = EC.

l.1(b). WC! = IBC! - IBFI = IAEI - ICEI = 21AEI - lAC! and IEC! =

lAC! - IAEI = lAC! - IBC!.

l.3(a).

n Pn qn rn

1 1 1 1

2 3 2 3/2 = 1.5

3 7 5 7/5 = 1.4

4 17 12 17/12 = 1.416666

5 41 29 41/29 = 1.413793

1.3(c). Clearly, rn > 1 for each n, so that from (b), rn+1 - rn and rn - rn-I have

opposite signs and

1

Irn+l - rnl < 41rn - rn-II·

Since r2 > 1 = rl, it follows that rl < r3 < r2. Suppose as an induction

hypothesis that

rl < r3 < ... < r2m-1 < r2m < ... < r2.

Then 0 < r2m - r2m+1 < r2m - r2m-1 and 0 < r2m+2 - r2m+1 < r2m - r2m+J.

so that r2m-1 < r2m+1 < r2m+2 < r2m. Let k, 1 be any positive integers. Then, for

m > k, I, r2k+1 < r2m-1 < r2m < r21.

l.3(d). Leta be the least upper bound (i.e., the smallest number at least as great as

all) of the numbers in the set {rl, r3, r5, ... , r2k+l, ... }. Then r2m-1 ::::: a ::::: r2m

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