If the squared difference of the zeros of the quadratic polynomial x^{2} + px + 45 is equal to 144 , find the value of p. (Ans: ± 18).
Ans: Let two zeros are α and β where α > β
According given condition
(α - β)^{2} = 144
Let p(x) = x^{2} + px + 45
α + β = -b/a = -p /1 = - p
αβ = c/a = 45/1 = 45
now (α - β)^{2} = 144
(α + β)^{2} - 4 αβ = 144
(-p)^{2} - 4 (45) = 144
Solving this we get p = ± 18