Solve the equation `x^2+p x+45=0.` it is given that the squared difference of its roots is equal to 144

Explanation

Let `alpha, beta` be the roots of the equation `x^(2)+px+45=0` and given that
`(alpha-beta^(2))=144`
`impliesp^(2)-4.1.45=144[ :' alpha-beta=(sqrt(D))/a]`
`impliesp^(2)=324`
`:.p=(+-18)`
On substituting `p=18` in the given equation we obtain
`x^(2)+18x+45=0`
`implies(x+3)(x+15)=0`
`impliesx=-3,5`
and substituting`p=-18` in the given equation we obtain
`x^(2)-18x+45=0`
`(x-3)(x-15)=0`
`impliesx=3,15` ltrbgt Hence the roots of the given are `(-3),(-15),3` and 15.