Question

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Q.Show that equation (a - 2)x² + (2 - b)x + (b - a) = 0 has equal roots. If 2a = b + 2.

Answer 1

Thus so

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Answer 2

Hi ,

Compare (a - 2 )x² + ( 2 - b )x + ( b - a ) = 0

with Ax² + Bx + C = 0 ,

A = ( a - 2 ),

B= 2 - b ,

C = b - a

And

2a = b + 2 ---( 1 )

Discreaminant ( D ) = B² - 4AC

= ( 2 - b )² - 4 × ( a - 2 ) × ( b - a )

= 4 - 4b + b² - 4 ( ab - a² - 2b + 2a )

= 4 - 4b + b² - 4ab + 4a² + 8b - 8a

= 4 + 4b + b² - 4ab + 4a² - 8a

= ( b + 2 )² - 2a ( 2b - 2a + 4 )

= ( b + 2 )² - ( b + 2 ) [ 2b - ( b + 2 ) + 4 ]

[ From ( 1 ) , ]

= ( b + 2 )² - ( b + 2 ) [ 2b - b - 2 + 4 ]

= ( b + 2 )² - ( b + 2 ) ( b + 2 )

= ( b + 2 )² - ( b + 2 )²

D = 0

Therefore ,

If D = 0 then given quadratic equation

has equal roots .

I hope this helps you.

: )