Question

If roots of $(a-2b+c)x^2+(b-2c+a)x+(c-2a+b)=0$ are equal, then :

• A. a = b = c
• B. a = c
• C. a = $2$b
• D. a = b

Answer 1

Answer: (D)

a = b

We have,

$\left(a-2b+c\right)x^2+\left(b-2c+a\right)x+\left(c-2a+b\right)=0$

Clearly $x=1$ is the row of equation.

Product of roots $x=1$ Sum of roots $=2$

$\frac{c-2a+b}{a-2b+c}=1$ $\frac{-\left(b-2c+a\right)}{a-2b+c}=2$

$-2a+b=a-b$ $2c-b-a=2c-4b+2a$

$3b=3a$ $3b=3a$

$b=a$ $b=a$

Hence,

Option $D$ is correct answer.