Question

if both roots of (b-c)x² + ( c - a) x + (a - b) = 0 are equal then prooved that a, b, c are in A.P .

Using rule 11th standard Please .

Answer 1

Hey there !

Thanks for the question !

Here's the answer !

We know that if a,b,c are in AP, their common difference remain same. So we get that, ( b - a ) = ( c - a ). If we prove this we can imply that a,b,c are in AP.

Given that both the roots are equal for the given quadratic equation. This implies that the discriminant of the equation is equal to zero.

Equation : ( b - c ) x² + ( c - a ) x + ( a - b ) = 0

A = ( b - c ) ; B = ( c - a ) ; C = ( a - b )

Discriminant = B² - 4AC = 0

=> ( c - a )² - 4 ( b - c ) ( a - b ) = 0

=> ( c² - 2ac + a² ) - 4 ( ab - b² - ac + bc ) = 0

=> c² - 2ac + a² - 4ab + 4b² + 4ac - 4bc = 0

=> a² + 4b² + c² - 4ab - 4bc + 2ac = 0 -> Equation 1

This is of the form of an identity:

( a + b + c )² = ( a² + b² + c² = 2ab + 2bc + 2ac )

So the Equation can be written as:

=> ( -a + 2b - c )² = 0 [ Simplifying using identity ]

Taking Square root on both sides we get,

=> -a + 2b - c = 0

=> - a + b + b - c = 0

=> b - a = c - b ( Transposing terms )

Hence we proved that the common differences are same.

Hence a,b,c are in AP.

Hence proved !

Answer 2

$\huge \pink{hello \: dear !!!}$



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Yeah ! Using rules of 11th Standard !


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