Question

solve for x: 4x^-4px+p^-q^=0, where possible and quality are integers.

Answer 1

let us consider p=q=1
4x^-4x=0
4x(x-1)=0
x=0,1

Answer 2

Answer:

Step-by-step explanation:

4x^2 - 4px + (p^2 - q^2) = 0

Here , a = 4 , b = -4p , c = p^2 - q^2

D = b^2 - 4ac

D = ( -4p )^2 - 4 (4) ( p^2 - q^2 )

D = 16 p^2 - 16 p^2 + 16 q^2

D = 16 q^2

x = - b _+√D / 2a

x = -( -4p ) +_ √16q^2 / 2(4)

x = 4p +_ 4q / 8

x = 4(p+_q) / 8

x = p+_q / 2

Either

x = p+q/2 OR x= p-q/2