If the roots of the equation (a – b)x2 + (b – c) x + (c – a) =

1.	If the roots of the equation (a – b)x2 + (b – c) x + (c – a) = 0 are equal, then A) 2b = a + c B) 2c = a + b C) 2a = b + c D) 2a = b – c
Answer» Correct option is (C) 2a = b + c For equal roots, we have Discriminant = 0 \(\Rightarrow(b-c)^2-4(a-b)(c-a)=0\) \(\Rightarrow(b^2+c^2-2bc)\) \(-(4ac-4a^2-4bc+4ab)=0\) \(\Rightarrow(b^2+c^2+4a^2+2bc-4ab-4ac)=0\) \(\Rightarrow b^2+c^2+(-2a)^2+2bc\) \(+2\times b\times -2a+2\times c\times -2a=0\) \(\Rightarrow(b+c-2a)^2=0\) \((\because a^2+b^2+c^2+2ab+2bc+2ac=(a+b+c)^2)\) \(\Rightarrow b+c-2a=0\) \(\Rightarrow\) 2a = b + c Correct option is C) 2a = b + c

If the roots of the equation (a – b)x2 + (b – c) x + (c – a) = 0 are equal, then A) 2b = a + c B) 2c = a + b C) 2a = b + c D) 2a = b – c

Answer»

Correct option is (C) 2a = b + c

For equal roots, we have

Discriminant = 0

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

\(\Rightarrow(b^2+c^2-2bc)\) \(-(4ac-4a^2-4bc+4ab)=0\)

\(\Rightarrow(b^2+c^2+4a^2+2bc-4ab-4ac)=0\)

\(\Rightarrow b^2+c^2+(-2a)^2+2bc\) \(+2\times b\times -2a+2\times c\times -2a=0\)

\(\Rightarrow(b+c-2a)^2=0\) \((\because a^2+b^2+c^2+2ab+2bc+2ac=(a+b+c)^2)\)

\(\Rightarrow b+c-2a=0\)

\(\Rightarrow\) 2a = b + c

Correct option is C) 2a = b + c