A.maths (quadratic equation and quadratic function)

The roots of x^2-px+q=0 are two consecutive integers
Let the roots be α and (α+1)

sum of the roots:
-(-p)÷1 = α + (α+1)
p = 2α + 1

product of the roots:
α(α+1) = q÷1
q = α(α+1)

∴LHS
=p^2-4q-1
=(2α+1)^2 - 4α(α+1) - 1
=4α^2 + 4α + 1 - 4α^2 - 4α - 1
=0
=RHS

Let the two roots be a and a+1
Sum of toot=p=2a+1
Product of root=q=a(a+1)
∴p^2-4q-1=4a^2+4a+1-4a^2-4a-1=0

Sum of root= -(-p)/1=p
Product of root = q
If the roots are two consecutive integers, then their difference must be 1
so let the root be a and b
a-b= Sqrt( (a+b)^2-4ab ) = Sqrt (p^2-4q)=1
Squaring both side
p^2-4q=1
so p^2-4q-1=0