Babette is thinking of a monic quadratic polynomial with real coefficients. She tells you that when she squares the roots of her quadratic, to form another monic quadratic with those squares as its roots, surprisingly, it is the same as the original. How many different quadratic polynomials could Babette have been thinking of?

This has something to do with sum of roots and product of roots of a quadratic.

greenplanet2050 Apr 19, 2024

#1**0 **

Let the roots of Babette's monic quadratic be r and s. Since the leading coefficient is 1, the quadratic can be written as:

p(x)=(x−r)(x−s)=x2−(r+s)x+rs.

We are given that squaring the roots gives another monic quadratic with those squares as its roots. In other words:

(x−r2)(x−s2)=x2−(r2+s2)x+r2s2

Expanding the left side gives:

x2−(r2+2rs+s2)x+r2s2

Equating the coefficients of the corresponding terms in both quadratics, we get the system of equations:

r+s=r2+s2

r2s2=rs (notice this simplifies to r2s2−rs=0)

From the first equation, we can rearrange to get r2+s2−r−s=0. Factoring this gives (r−1)(s−1)=0. This means either r=1 or s=1.

We can consider two cases:

Case 1: r=1

Substituting r=1 into the second equation gives s2−s=0, which factors as s(s−1)=0. This means either s=0 or s=1. However, if s=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

Therefore, in this case, s=1 and the quadratic is simply x2−2x+1=(x−1)2. There is only 1 quadratic possible in this case.

Case 2: s=1

Substituting s=1 into the second equation gives r2−r=0, which factors as r(r−1)=0. This means either r=0 or r=1. However, if r=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

Therefore, in this case, r=1 and the quadratic is again simply x2−2x+1=(x−1)2. There is only 1 quadratic possible in this case.

Since both cases lead to only one possible quadratic, Babette could have been thinking of at most 2 different monic quadratics.

bader Apr 19, 2024