Question

If the squared difference of the series of the quadratic polynomial f(x) = x²+px+45 is equal to 144, then find the value of p.

Answer 1

Hey

Here is your answer

The given quadratic polynomial is f(x) = x2 + px + 45.

Let ∝ and β be the zeroes of the given quadratic polynomial.

∴∝ + β = – p and ∝β = 45 ...(1)

Given, (∝ – β)2 = 144

∴ (∝ + β)2 – 4∝β = 144

⇒ (– p)2 – 4 × 45 = 144 [Using (1)]

⇒ p2 – 180 = 144

⇒ p2 = 144 + 180 = 324

p=+/-√324=+/-18

Thus, the value of p is ± 18.

Hope it helps you!