Question

(b-c) x square +(c-a) x +(a+b) =0 has equal roots. Show that 2b=a+c​

Answer 1

Answer:

Step-by-step explanation:

b-c)x²+(c-a)x+(a-b)=0

Comparing with quadratic equation

Ax²+Bx+C=0

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

Discriminate when roots are equal

D=B²-4AC=0

D=(c-a)²−4(b-c)(a-b)=0

D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0

D=c²+a²−2ac-4ab+4ac+4b²-4bc=0

c²+a²+2ac-4b(a+c)+4b²=0

(a+c)²-4b(a+c)+4b²=0

[(a+c)-2b]²=0

a+c=2b

Answer 2

Answer:

Step-by-step explanation:

( b - c ) x² + ( c - a ) x + ( a + b ) = 0 has equal roots ,

then ,

b² - 4ac = 0

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

( c - a )² - 4ba + 4b² + 4ac + 4bc = 0

( c - a )² + 4ac - 4b ( a + c )  + 4b² = 0

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

( a + c - 2b )² = 0

a + c - 2b = 0

a + c = 2b

2b = a + c  [ Proved ]