Question

solve p2x2+(p2-q2)x-q2=0

Answer 1

Hi ,


Given quadratic equation is 


p²x² + ( p² - q² )x - q² = 0

dividing each term with p² ,

p²x² / p² + [ ( p² - q² ) / p² ] x - q² / p² = 0

⇒ x²  + [ 1 - q² / p² ] x - q² / p² = 0

 To find the roots split the middle term

⇒     x² + x - ( q² / p² ) x - q² / p² = 0

⇒  x ( x + 1 ) - ( q² / p² ) ( x + 1) = 0

⇒                ( x+ 1 ) ( x - q² /p² ) = 0

Therefore,

                 x+1 =0  or  x - q² / p² =0

            x= -1 or       x  = q² / p²

I hope this helps you.

****

 

Answer 2

Answer:

p2x2+(p2-q2)x-q2 =0

p2x+p2x-q2x-q2 =0

a = p2

b = p2-q2

c = -q2

using quadratic formula,

D = -b+-√(b2-4ac)/2a

= -(p2-q2)+-√[(p2-q2)^2-4×p2×q2]/2p^2

= -p2+q2+-√[p4-q4-2p2q2+4p2q2]/2p^2

= -p2+q2+-√(p4-q4+2p^2q^2)/2p^2

= -p2+q2+-p2-q2/2p^2

= -p2+q2-p2-q2/2p^2 , -p2+q2+p2+q2/2p^2

= -2p^2/2p^2 , 2q^2/2p^2

= -1 , q2/p2