Question

If the roots of the equation a-b x2 +b-c x +c-a=0are equal prove that 2a =b+c

Answer 1

Given real roots

∆=0

B^2-4AC=0

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

By solving we get,

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

2a=b+c

Answer 2

Answer:

Step-by-step explanation:

Hi,

We know that,

If the quadratic equation=ax²+bx+c=0 whose roots are equal then it's deteminant is equal to zero.

=>(a-b)x²+(b-c)x+(c-a)=0

Deteminant =0

=>(b-c)² -4(a-b)(c-a)=0

=>b²+c²-2bc-4ac+4a²+4bc-4ab=0

=>b²+c²+4a²+4bc-4ac-4ab=0

=>b²+c²(-2a)²+2bc+2c(-2a)+2(-2a)b=0

=>(b+c-2a)²=0

=>b+c-2a=0

Therefore,

b+c=2a

Hence proved.