Question

if the squared difference of the zeroes of the quadratic polynomail f(x) is equal to x² +px+45 is equal to 144 , find the value of p.

Answer 1

Answer:

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.