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) + 45 = 0`
Then, `alpha + beta = -p` (1)
`alpha beta = 45` (2)
it is given that
`(alpha = beta)^(2) = 144`
or `(alpha + beta)^(2) - a alpha beta = 144`
or `p^(2) - 4xx 45 = 144`
or `p^(2)324`
or `P = pm 18`
Substitueing = 18 in the given equation, we obtain
`x^(2) + 18x 45 = 0`
or `(x+ 3) (x + 15) = 0`
or x = -3, -15
Substituting p = - 18 in the given equation, we obtain
`x^(2) + 18x + 45 = 0`
`rArr (x - 3) (x - 15) = 0`
`rArr x = 3, 15`
Hence, the roots of the given equation are -3, -15 or 3, 15.