If the roots of the equation (a-b)x²+(b-c)x+(c-a)=0 are equal, prove that 2a=b+c
Answers
Answered by
12
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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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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LegendaryGenius:
thank you
Answered by
12
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 .
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 .
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