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一條A maths(Quad. Function)
If the equations ax^2 + bx + c = 0 and px^2 + qx + r=0 have a root in common, show that (br - cq)(aq - bp) = (cp - ar)^2
最佳解答:
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
其他解答:
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
一條A maths(Quad. Function)
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發問:If the equations ax^2 + bx + c = 0 and px^2 + qx + r=0 have a root in common, show that (br - cq)(aq - bp) = (cp - ar)^2
最佳解答:
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
其他解答:
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
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