Algebra questions!!

1. Using Vieta's formulas, we know that A + B = 5 and AB = 3. We want to find a quadratic with roots A^2 and B^2. 

We know that (x - A^2)(x - B^2) = 0 has roots A^2 and B^2. Expanding this equation, we get:

x^2 - (A^2 + B^2) x + A^2B^2 = 0

Now we need to express A^2 + B^2 and A^2B^2 in terms of A and B. We know that:

(A + B)^2 = A^2 + 2AB + B^2

(A - B)^2 = A^2 - 2AB + B^2

Adding these two equations gives:

2A^2 + 2B^2 = 2(A^2 + B^2) + 2AB

Simplifying, we get:

A^2 + B^2 = (A + B)^2 - 2AB
A^2 + B^2 = 25 - 2(3)
A^2 + B^2 = 19

Using the fact that AB = 3, we can substitute into the quadratic we derived earlier to get:

x^2 - 19x + 9 = 0

Therefore, the quadratic whose roots are A^2 and B^2 is x^2 - 19x + 9.

2. We have A - B = 4 and A^3 - B^3 = 52. 

(a) To find all possible values of AB, we can use the identity:

A^3 - B^3 = (A - B)(A^2 + AB + B^2)

Substituting in A - B = 4 and simplifying, we get:

52 = 4(A^2 + AB + B^2)
13 = A^2 + AB + B^2

Using the fact that AB = (A - B)^2 - A^2 - B^2, we can substitute into the equation above to get:

13 = 2A^2 + 2B^2 - 8A
13 = 2(A - 2)^2 + 2B^2 - 17

Rearranging and simplifying, we get:

2B^2 = -2(A - 2)^2 + 30
B^2 = -(A - 2)^2 + 15

Since B^2 is non-negative for all real numbers, we know that -(A - 2)^2 + 15 ≥ 0. Solving for A, we get:

(A - 2)^2 ≤ 15

Therefore, -√15 + 2 ≤ A ≤ √15 + 2.

Now we canuse this range of values for A to find the corresponding range of values for AB. We know that AB = 3 + B(A - B), so substituting in A - B = 4, we get:

AB = 3 + 4B

Since B can take any real value, the range of possible values for AB is (-∞, ∞).

(b) To find all possible values of A + B, we can add A - B = 4 to A^3 - B^3 = 52 to get:

A^3 + 3A^2B + 3AB^2 + B^3 = 56

Using the fact that A^3 - B^3 = 52 and A - B = 4, we can substitute and simplify to get:

4A^2 + 12AB + 4B^2 = 56
A^2 + 3AB + B^2 = 14

Using the fact that A + B = (A - B) + 2B = 4 + 2B, we can substitute and simplify to get:

A^2 + 3AB + B^2 = (A + B)^2 - AB = 14
(A + B)^2 = 14 + AB

Since AB can take any real value, the range of possible values for (A + B)^2 is [14, ∞). Taking the square root of both sidesand considering both positive and negative values, we get:

A + B = ±√(14 + AB)

Therefore, the possible values of A + B are A + B = √(14 + AB) and A + B = -√(14 + AB).

(c) Using the expression we derived in part (a), we know that:

B^2 = -(A - 2)^2 + 15

Since B^2 is non-negative for all real numbers, we know that -(A - 2)^2 + 15 ≥ 0. Solving for A, we get:

(A - 2)^2 ≤ 15

Therefore, -√15 + 2 ≤ A ≤ √15 + 2. 

Using the fact that A - B = 4, we can solve for B in terms of A to get:

B = A - 4

Substituting the range of values for A, we get:

-√15 - 2 ≤ B ≤ √15 - 2

Therefore, the possible values of A and B are:

-√15 + 2 ≤ A ≤ √15 + 2, and -√15 - 2 ≤ B ≤ √15 - 2.