Find the value of p if the roots of the quadratic equation 4x^2 -(5p+1)x+5p=0 differ by 1.

We'll apply Viete's relations to determine the value of
p.

Viete's relations link the coefficients of a polynomial
to its roots.

For instance, if the polynomial is a
quadratic, ax^2 + bx + c = 0 Viete's relations are the
followings:

x1 + x2 =
-b/a

x1*x2 = c/a

x1 and x2 are
the roots of the quadratic

a,b,c are the
coefficients:

We'll identify the
coefficients:

a = 4, b = -(5p+1) , c =
5p

Since the roots have to be different from 1, we'll get
the constraints:

The sum S = x1 + x2 has to be different
from 2 and the product P = x1*x2 has to be different from
1:

x1 + x2 = (5p+1)/4

(5p+1)/4
is different from 2.

We'll put (5p+1)/4 =
2

5p + 1 = 8

5p =
8-1

5p = 7 => p =
7/5

The product x1*x2 = 5p/4 has to be different from
1:

5p/4 = 1 => 5p = 4 => p =
4/5

The values of p are all real numbers,
except the following: {4/5 ; 7/5}.