一條A maths(Quad. Function)

Rearrange the two equations as x2 + ( b / a ) x + ( c / a ) = 0 and x2 + ( q / p ) x + ( r / p ) = 0

Then by subtraction,

( b / a – q / p ) x + ( c / a – r / p ) = 0

( b / a – q / p ) x = r / p – c / a

x = ( ar – cp ) / ( bp – aq )

Then put it back into the equation, ( since there's a common root )

( ar – cp )2 / ( bp – aq )2 + ( b / a ) ( ar – cp ) / ( bp – aq ) + ( c / a ) = 0

( ar – cp )2 + ( b / a )( ar – cp )( bp – aq ) + ( c / a )( bp – aq )2 = 0

( ar – cq )2 + ( bp – aq ){ ( b / a ) ( ar – cp ) + ( c / a )( bp – aq ) } = 0

( ar – cq )2 + ( bp – aq )( abr – caq ) / a = 0

( ar – cq )2 = -( br – cq )( bp – aq )

Therefore (br – cq )( aq – bp ) = ( cq – ar )2

as above, but for the line
x = ( ar – cp ) / ( bp – aq )
case 1: bp - aq <> 0
follow

case 2: bp - aq = 0
=> ra - pc = 0
=> (br - cq)(aq - bp) = (cp - ar)^2