# Proposed Problem

Discussion in 'Mathematics' started by AVDBA, Sep 17, 2019.

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1. Let r and s be the roots of the equation f(x) = a + x^2 + (a - 15)x = 0. Given that r and s are integers, find the set of possible values for a.

2. Are you asking for solutions or comments whether this is a suitable problem to set for your students? If the latter it depends what stage the students are at.

3. Thanks gainly. I am providing these latter proposed problems, since I think these are two examples that explore a process of "working with the concepts". For these two cases, they are related to interconnecting concepts regarding factorability. They are not intended to just reach their solutions. For reaching a solution, one needs to work with the concepts throughout the solution. They are examples of "exploring the concepts throughout". The related concepts are, for these specifically, their tags. This problem here, specifically, requires factoring a polynomial using roots, so reaching the Girard relations, as well as the analysis to impose the condition that the solution must obey the necessity required by the set as stated by the problem's statement.

4. Hints for working throughout the solution:

1 - To write the Girard relations relating the roots r and s, which provides two relations.

2 - By eliminating a from the previous relations, one finds a factorable result that turns out to have an important property.

3 - To use the Girard relations to write a as depending on one root. Keeping in mind that factorability is important.

4 - To test all values that fall into the required set (controlled test).

5. If I was writing hints to help my brighter students with this problem I would use much simpler terms. I think they would be completely baffled by your hints.

6. 0,7,27,34

7. Also 9,25

So 0,7,9,25.27,34

8. Thanks gainly. It was intended to guide through a solution.

Writing the quadratic equation under the factorable form:

(x - r)(x - s) = 0

one reaches:

x^2 -sx -rx + sr = 0

or:

sr + x^2 - (r + s)x = 0

giving the Girard relations:

rs = a

(r + s) = 15 - a

rs + r + s = 15

or:

r(s + 1) + s + 1 - 1 = 15

giving:

(r + 1)(s + 1) = 16

We conclude that r + 1 divides 16. Hence, the values of r + 1 are
within the set of divisors of 16: {-16, -8, -4, -2, -1, 1, 2, 4, 8, 16}.
The possible values for r are within the set R = {-17, -9, -5, -3,-2, 0,
1, 3, 7, 15}, thence. Now, using the Girard relations:

r + a/r = 15 - a

or:

a(1/r + 1) = 15 - r => a = r(15 - r)/(1 + r). By using the set R for the
possible values of r, one obtains that that the possible values for a
are within the set A = {0, 7, 9, 25, 27, 34}.