Question

urgent! !!!!!! for 25 points
if the squared difference of the zeros of the quadratic polynomial f (x)=x square+px+45 is equal to 144, find value of p

Answer 1

f(x)=x²+PX+45
a =1,b=p,c=45
these are the values now we know that
alpha+beta=-b\a
=-p.
alpha×beta=c\a
45.
they asked difference of their squared zeros which means
alpha²-beta²=144.
after this ull be able to do.

Answer 2

f(x)=x²+px+45
Let α,β be the roots of the quadratic equation
a=1,b=p and c=45

Now, we know that,
α+β=-b/a=-p/1=-p
αβ=c/a=45

given,
       (α-β)²=144
        α²+β²-2αβ=144
        -p-2(45)=144
        -p-90=144
         -p=144+90
         ∴p=234