Algebra mathmatich question

Question

The equation x2 + px +q = 0 is given. Find parameters p and q such that the difference of roots of the given equation is 5 and that the difference of cubes of roots is 35.

Answer 1

x^2 + px + q = 0

p=-1, q=-6

x^2-x -6

(x-3)(x+2)=0 roots are x=-2 and x=3. 3- -2 = 5, 3^3 -(-2)^3 = 27+8=35

x = -p/2 + or - (1/2)sqr(p^2 -4q)

(1/2)sqr(p^2-4q) = -(1/2)sqr(p^2-4q) + 5

sqr(p^2 -4q) = 5

p^2 -4q = 25

[-p/2 +(1/2)sqr(p^2-4q)]^3 -[-p/2 - (1/2)sqr(p^2-4q)]^3 =35

(-p/2 +(1/2)(5))^3 -(-p/2-(1/2)(5))^3 =35

(-p+125)^3 -(-p-125)^3 =35(2)^3

3p^2(5) + 125 = 140

3p^2(5) =15

p^2 =1

p=-1

sqr(1-4q)=5

q=-6

there may also be a 2nd solution

p=+1

q=-6

x^2+x-6 = (x+3)(x-2)

roots x=-3 and x=2

their difference = 5

difference of cubes = 2^3 - (-3)^3 = 8--27 = 35