If p, q are roots of the equation ax^2 + bx + c = 0, then find the equation whose roots are (2p + 3q) and (3p + 2q)

Given that p and q are roots of equation ax^2 + bx + c = 0

So, Sum of roots = -b/a
=> p + q = -b/a                  - (Equation 1)

Product of roots = c/a
=> pq = c/a                        - (Equation 2)

Now, we have to find the equation whose roots are (2α + 3) and (3α + 2).

Sum of roots of that equation will be
(2p + 3q) + (3p + 2q)
= (5p + 5q)
= 5(p + q)


Putting the value of (p + q) from Equation 1,
= 5(-b/a) = -5b/a
 
Product of roots of that equation will be

(2p + 3q) (3p + 2q)
= 6p^2 + 9pq + 4pq + 6q^2
= 6 (p^2 + q^2) + 13pq 
= 6 [(p + q)^2 - 2pq)] + 13pq
= 6(p + q)^2 - 12pq + 13pq
= 6(p + q)^2 + pq

Substituting the values of (p + q) and pq from Equation 1 and 2, 

= 6(-b/a)^2 + c/a
= 6(b^2/a^2) + c/a
= (6b^2 + ac) / a^2

We know that an equation whose sum of roots is m and product of roots is n can be written as

x^2 - mx + n = 0

So, the equation we want is

x^2 - (-5b/a) + [(6b^2 + ac) / a^2] = 0

Multiplying both the sides with a^2,

(a^2)x^2 + 5abx + 6b^2 + ac = 0 is the final answer.