Question

If the difference of the squares of the zeroes of the quadratic equation x2+px+45 is 144 find p

Answer 1

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.

Answer 2

Hey Joddar,

Let a,b are the zeroes of

p(x)=x^2+px+45.

=>a+b=-p/1=-p, ab=45/1=45

Also,|a-b|^2=144.(given).

:(a+b)^2-4ab=|a-b|^2

=>(-p)^2-4(45)=144

=>p^2=144+180=324

=>p=√324 or -√324

=>p=18 or p=-18.

Hope it helps ✌✌