Question

a(b-c)+b(c-a)+c(a-b) has equal roots .
Prove that 1/a+1/b=2/c

Answer 1

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Answer 2

Answer:

Proved

Step-by-step explanation:

a(b-c)+b(c-a)+c(a-b) has equal roots .

Prove that 1/a+1/b=2/c

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

so

(b(c-a))² = 4a(b-c)c(a-b)

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

=> b² (c² + a² - 2ac) + 4acb² = 4acb(a+c) - 4(ac)²

=> b² (c² + a² - 2ac + 4ac) = 4acb(a+c) - (2ac)²

=> b² (c² + a² + 2ac) - 4acb(a+c) + (2ac)² = 0

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

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

=>  (b(c+a) - 2ac)² = 0

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

=> bc + ba = 2ac

Dividing by abc both sides

=> 1/a  + 1/c  = 2/b

QED