answer:

Step-by-step explanation:

Let p ( x ) = a²x² - ( a ²b² + 1 ) x + b² = 0

On comparing the given equation with the quadratic equation ax² + bx + c = 0,

we have a = a², b = - ( a²b² + 1 ) and c = b².

d = b² - 4ac

= [ - ( a²b² + 1 ) ]² - 4 ( a²b² )

= ( a²b² )² + 1 + 2 ( a²b² ) ( 1 ) - 4a²b²

= ( a²b² )² + 1 + 2a²b² - 4a²b²

= ( a²b² ) + 1 - 2a²b²

= ( a²b² - 1 )²

By using quadratic formula,

x = - b ± √ b² - 4ac / 2a

= - [ - ( a²b² + 1 ) ] ± √ ( a²b² - 1 )² / 2a²

= ( a²b² + 1 ) ± ( a²b² - 1 ) / 2a²

= ( a²b² + 1 + a²b² - 1 ) / 2a² or [ a²b² + 1 - ( a²b² - 1 ) ]/ 2a²

= 2a²b² / 2a² or ( a²b² + 1 - a²b² + 1 ) / 2a²

= b² or 2/2a²

= b² or 1/ a²

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