Question

If the roots of the equation (a-b)x²+(b-c)x+(c-a)=0 are equal, prove that 2a=b+c

Answer 1

Hello here is your answer by Sujeet yaduvanshi ☝☝☝☝☝☝☝☝☝☝

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

Applying Discriminant

D=b²-4ac

D=(b-c)²-4(a-b)(c-a)


Also,

GIVEN
For real and Equal roots ,We must be have

D=0

then


(b-c)²-4(a-b)(c-a)=0

4a²+b²+c²-4ab+2bc-4ac=0

(-2a)²+b²+c²+2(-2a)b+2bc+2c(-2a)=0

This is the form of (a+b+c)²


(-2a+b+c)²=0


-2a+b+c=0

b+ c=2a


means...

2a=b+c



that's all


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Answer 2

Hey user

Here is your answer :-

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

Comparing the equation with ax² + bx + c = 0

Here a = ( a - b )
b = ( b - c )
c = ( c - a )

Given that the equation has equal roots

So descriminant should be 0 .

But

D = b² - 4ac

0 = ( b - c )² - 4( a - b )( c - a )

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

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

0 = 4a² + b² + c² + 2bc - 4ac + 4ab

0 = ( -2a )² + b² + c² + 2bc - 2 ( -2a )c + 2( -2a )b

0 = ( -2a + b + c )²

0 = - 2a + b + c

2a = b + c .