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Roots of the equation is given by -b +-√b²-4ac/2a -(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