Question

prove that (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 have equal roots if a=b=c.​

Answer 1

Answer:

Step-by-step explanation:

→given if a=b=c

→so,

→[x-a].[x-b]+[x-b][x-c]+[x-c][x-a]=0

⇒3x²-[ 2{a+b+c} ] + ab+bc+ac=0

→as a=b=c

⇒3x²-[2{3a}]x + a.a+a.a+a.a=0

⇒3x²-6ax + 3a²=0

⇒x²-2ax+a²=0

→as we have formula -b±√b²-4ac÷2a

⇒2a±√4a²-4a²/2

⇒2a±√0/2

⇒2a/2

⇒a