If the squared difference of the zeros of the quadratic polynomial x² + 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) = x² + 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