Question

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

Answer 1

SOLUTION :  

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

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

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

Discriminant(D) = b² -2ac  

D = (c - a)² - 4(b - c)(a - b)  

D = c² + a² - 2ac – 4(ab - b² - ac + cb)  

[(a - b)² = a² + b² - 2ab]

D = c² + a² - 2ac – 4ab + 4b² + 4ac - 4cb  

D = c² + a² - 2ac + 4ac – 4ab + 4b² - 4cb  

D = c² + a² + 2ac  – 4ab + 4b² - 4cb  

D = a² + 4b² + c² + 2ac  - 4ab - 4cb  

As We know that , (a + c - 2b)² = a² + 4b² + c² + 2ac  - 4ab - 4cb  

D = (a + c - 2b)²  

Real and equal root are Given  

Discriminant(D) = 0

0 = (a + c - 2b)²

Put Square root on both sides  

0 = √(a + c - 2b)²

0 = (a + c - 2b)

a + c =  2b

Hence , 2b = a + C  

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

The roots are equal .

Let one root be p .

Hence sum of roots will be p + p = 2 p

Now see the polynomial .

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

Compare with a x² + bx + c = 0 we get :

a = b - c

b = c - a

c = a - b

Sum of roots = - b / a

2 p = - b / a

= > 2 p = - ( c - a ) / ( b - c )

= > 2 p = ( a - c ) / ( b - c )

= > p = ( a - c ) / 2 ( b - c ) ........( 1 )

Product of roots = c / a

= > p × p

= > p² = ( a - b ) / ( b - c ) .............( 2 )

Square ( 1 ) to get :

p² = ( a - c )² / 4 ( b - c )² ...........( 3 )

From ( 3 ) and ( 2 ) we have :

( a - c )² / 4 ( b - c )² = ( a - b ) / ( b - c )

= > ( a - c )² = 4 ( a - b )( b - c )

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

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

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

= > ( a - 2 b + c )² = 0

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

= > a + c= 2 b

Hence proved