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

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### 1 Answer

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}.**