If the roots of the equation (a – b) x2 + (b – c) + (c – a) = 0 are equal, prove that b + c = 2a.
As the equation is in the form Ax2 + Bx + C = 0 with non-zero A.
In which,
A = a - b
B = b - c
C = c - a
And we know that if the roots of a equation are equal then we have
Discriminant, D = 0
Where, D = b2 - 4ac
⇒ b2 - 4ac = 0
⇒ (b - c)2 - 4(a - b)(c - a) = 0
⇒ (b - c)2 + 4(a - b)(a - c) = 0
⇒ b2 + c2 - 2bc + 4(a2 - ac - ab + bc) = 0
⇒ b2 + c2 - 2bc + 4a2 - 4ac - 4ab + 4bc = 0
⇒ 4a2+ b2 + c2 - 4ac + 2bc - 4ac = 0
⇒ (-2a)2+ b2 + c2 + 2(-2a)c + 2bc + 2(-2a)c = 0
⇒ (-2a + b + c)2 = 0
[using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2za ]
⇒ -2a + b + c = 0
⇒ b + c = 2a
Hence Proved.