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If the roots of the quadratic equation (a-b)x²+(b-c)x+(c-a)=0 are equal, prove that 2a=b+c.

I need quick answer

1
by 201301

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I need quick answer

by 201301

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-(b-c)+√(b-c)²-4(c-a)(a-b)/2(a-b)=-(b-c)-√(b-c)²-4(c-a)(a-b) both roots are equal

2√{(b-c)²-4(c-a)(a-b)}/2(a-b)=0

squaring both side

(b-c)²-4(c-a)(a-b)=0

b²+c²-2bc=4(ac-bc-a²-ab)

b²+c²-2bc= 4ac-4bc-4a²-4ab

b²+c²+2bc=4ac-4ab-4a²

(b+c)²=4(ac-ab -a²) ac=ab because both roots are equal

(b+c)²=-4a²

taking root on both side to make a² +ve

sp √(b+c)²=√4a²

hence b+c=2a