Question

If squared difference of zeroes of the quadratic polynomial f(x) =x^2+px+45 is equal to 144, find the value of p

Answer 1

★ QUADRATIC EQUATIONS ★

Let α,β are the roots of the quadratic polynomial f(x) = x2+px+45 then 

 α + β = -p ---------(1) and  αβ = 45

Given  (α - β)2 = 144 

∴ (α + β)2 – 4αβ = 144

⇒ (– p)2 – 4 × 45 = 144               [Using (1)]

⇒ p 2 – 180 = 144

⇒ p 2 = 144 + 180 = 324

Thus, the value of p is ± 18.