Q1. If the roots of the equation (x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0 are equal , then a) a+B+C=0 b) a+b(omega)+c(omega)square=0 c) a-b+c=0 d) none ofd these

							
Given equation is
(x - b)(x - c) + (x - c)(x - a) + (x - a)(x - b) = 0
On simplification, equation changes to
x2 - (b + c)x + bc + x2 - (c + a)x + ca + x2 - (a +b)x + ab = 0
3x2 - (2a + 2b + 2c)x + (ab + bc + ca) = 0
3x2 - 2(a + b + c)x + (ab + bc + ca) = 0
On comparing the above equation with Ax2 + Bx + C = 0,
A = 3, B = -2(a + b + c), C = (ab + bc + ca)
Since roots are equal
Discriminant D = 0
      B2 - 4AC = 0
[-2(a + b + c)]2 - 4(3)(ab + bc + ca) = 0
4(a + b+ c)2 -12(ab + bc + ca) = 0
4(a2 + b2 + c2 + 2ab + 2bc + 2ca) - 12ab -12bc -12ca = 0
4a2 + 4b2 + 4c2 + 8ab + 8bc + 8ca - 12ab -12bc -12ca = 0
4a2 + 4b2 + 4c2 - 4ab -4bc -4ca = 0
4(a2 + b2 + c2 - ab -bc -ca) = 0
Dividing both sides by 4
a2 + b2 + c2 - ab -bc -ca= 0
 
Hence answer is option D - None of these
 
						

							
(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0
On simplification, we get
3x^2 -2(a+b+c)x + (ab+bc+ca) = 0.
Roots are equal implying that D = 0.
i.e 4(a+b+c)^2 -4(3)(ab+bc+ca) = 0
or, 4[a^2 + b^2 +c^2 +2(ab+bc+ca) -3(ab+bc+ca)]=0
or, 4[a^2 + b^2 + c^2 -ab-bc-ca]=0
or, 2[2a^2 + 2b^2 +2c^2 - 2ab -2bc-2ca]=0
or, (a^2 + b^2 -2ab) + (b^2 + c^2 -2bc) + (c^2 + a^2 -2ac)=0
i.e (a-b)^2 + (b-c)^2 + (c-a)^2 = 0
This is true only if all three terms are zero.
Hence a=b=c.
Now option B becomes a( 1+omega + omega(squared) ) = a*0 = 0.
Hence option B is correct.