Question

if the roots of the eq (b-c) x2 + (c-a)x+(a-b)=0 are equal prove that 2a=a+c

Answer 1

Hi ,

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

compare given quadratic equation with

Ax² + Bx + C = 0 ,

A = b - c ;

B = c - a ;

C = a - b ;

it is given that roots of the given equation are

equal.

discreaminant = 0

B² - 4AC = 0

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

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

c² + a² + 4b² + 2ac - 4ab - 4bc = 0

c² + a² + (- 2b)²+2×ca + 2 ×a×(-2b ) +2×(-2b)c=0

( c + a - 2b )² = 0

Therefore ,

c + a - 2b = 0

c + a = 2b

Hence proved .

I hope this helps you.

:)

Answer 2

If the roots of quadratic equation are equal, discriminant is 0 .

b² - 4ac = 0 

(c-a) ² - 4 ( b-c) (a - b ) = 0

c²+a²-2ac -4 { ab-b²-ac+bc} = 0

c²+a²-2ac-4ab+4b²+4ac-4bc= 0 

a²+c²+(-2b)²+2(a)(c) +2(-2b)(a) +2(-2n)(c) = 0

( a + c -2b) ² = 0 

a+ c -2b = 0

a+c=2b .

Hope helped!